Roots And Parabolas A Comprehensive Guide To Quadratic Functions
Hey guys! Today, we're diving deep into the fascinating world of quadratic functions and their graphical representations parabolas. We'll be tackling a series of quadratic equations, finding their roots (where the parabola intersects the x-axis), and then sketching those beautiful curves. So, buckle up and let's get started!
Understanding Quadratic Functions and Parabolas
Before we jump into solving, let's quickly recap the basics. A quadratic function is a polynomial function of degree two, generally expressed in the form f(x) = ax² + bx + c, where a, b, and c are constants and 'a' is not equal to zero. The graph of a quadratic function is a parabola, a symmetrical U-shaped curve. The coefficient 'a' plays a crucial role in determining the parabola's shape: if 'a' is positive, the parabola opens upwards (a smiley face!), and if 'a' is negative, it opens downwards (a frowny face!). The roots of the quadratic function are the x-values where the parabola intersects the x-axis, also known as the zeros of the function. Finding these roots is essential for sketching the parabola accurately. We'll use the quadratic formula, a powerful tool for finding roots, as well as factoring techniques where applicable. Understanding the discriminant (b² - 4ac) within the quadratic formula is key because it tells us the nature of the roots: a positive discriminant indicates two distinct real roots, a zero discriminant indicates one real root (a repeated root), and a negative discriminant indicates two complex roots (meaning the parabola doesn't intersect the x-axis). The vertex of the parabola, the point where the curve changes direction, is another critical feature. Its x-coordinate can be found using the formula -b/2a, and substituting this value back into the function gives us the y-coordinate. This vertex represents the minimum point if the parabola opens upwards and the maximum point if it opens downwards. By finding the roots, the vertex, and knowing the direction the parabola opens, we can sketch an accurate representation of the quadratic function. Let's put these concepts into practice with our examples.
a) f(x) = -x² + 4x - 3
Alright, let's kick things off with our first quadratic function: f(x) = -x² + 4x - 3. Our main goal here is to determine the roots of this function and then represent it as a parabola. The roots, as we know, are the x-values where the function equals zero. So, we need to solve the equation -x² + 4x - 3 = 0. To do this effectively, we can use the quadratic formula. But before we jump into that, let's see if we can factor this quadratic. Factoring, when possible, is often a quicker and cleaner method. We're looking for two numbers that multiply to -3 * -1 = 3 (the product of the leading coefficient and the constant term) and add up to 4 (the coefficient of the x term). Those numbers are 3 and 1! So, we can rewrite the equation as -(x² - 4x + 3) = -(x - 3)(x - 1) = 0. Setting each factor to zero, we get x - 3 = 0 or x - 1 = 0, which gives us the roots x = 3 and x = 1. Excellent! We've found our roots. Now, to sketch the parabola, we need a bit more information. Since the coefficient of the x² term is negative (-1), the parabola opens downwards. Next, let's find the vertex. The x-coordinate of the vertex is given by -b/2a, which in our case is -4 / (2 * -1) = 2. Plugging x = 2 back into the function, we get f(2) = -(2)² + 4(2) - 3 = -4 + 8 - 3 = 1. So, the vertex is at the point (2, 1). With the roots (1, 0) and (3, 0), the vertex (2, 1), and the fact that the parabola opens downwards, we can sketch a pretty accurate parabola. It's a downward-facing curve intersecting the x-axis at x = 1 and x = 3, with its highest point at the vertex (2, 1). This gives us a clear picture of the function's behavior and its graphical representation.
b) f(x) = 3x² - 12x + 5
Now, let's tackle the function f(x) = 3x² - 12x + 5. Again, our primary objective is to find the roots and then graph the parabola. Unlike the previous example, factoring this quadratic might not be as straightforward. So, let's bring out the big guns the quadratic formula! The quadratic formula states that for an equation of the form ax² + bx + c = 0, the roots are given by x = (-b ± √(b² - 4ac)) / 2a. In our case, a = 3, b = -12, and c = 5. Plugging these values into the formula, we get x = (12 ± √((-12)² - 4 * 3 * 5)) / (2 * 3) = (12 ± √(144 - 60)) / 6 = (12 ± √84) / 6. Simplifying the square root, √84 = √(4 * 21) = 2√21. So, our roots are x = (12 ± 2√21) / 6. We can further simplify this by dividing both the numerator and denominator by 2, giving us x = (6 ± √21) / 3. Thus, we have two distinct real roots: x₁ = (6 + √21) / 3 and x₂ = (6 - √21) / 3. To get a better sense of where these roots lie on the x-axis, we can approximate √21 to be around 4.58. This gives us x₁ ≈ (6 + 4.58) / 3 ≈ 3.53 and x₂ ≈ (6 - 4.58) / 3 ≈ 0.47. Now, let's find the vertex. The x-coordinate of the vertex is -b/2a = -(-12) / (2 * 3) = 12 / 6 = 2. Plugging x = 2 back into the function, we get f(2) = 3(2)² - 12(2) + 5 = 12 - 24 + 5 = -7. So, the vertex is at the point (2, -7). Since the coefficient of the x² term is positive (3), the parabola opens upwards. With the roots approximately at x = 0.47 and x = 3.53, the vertex at (2, -7), and the upward-opening nature of the parabola, we can sketch its graph. It's a U-shaped curve intersecting the x-axis at our calculated roots, with its lowest point at the vertex. This gives us a solid understanding of the function's behavior and its visual representation.
c) f(x) = x² - 7x + 12
Let's move on to the function f(x) = x² - 7x + 12. As before, our mission is to determine the roots and then sketch the parabola. This time, let's try factoring first. We need to find two numbers that multiply to 12 and add up to -7. Thinking about the factors of 12, we quickly realize that -3 and -4 fit the bill perfectly! (-3) * (-4) = 12 and (-3) + (-4) = -7. So, we can factor the quadratic as (x - 3)(x - 4) = 0. Setting each factor to zero, we get x - 3 = 0 or x - 4 = 0, which gives us the roots x = 3 and x = 4. Awesome! We've easily found the roots. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (1), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate, which is -(-7) / (2 * 1) = 7 / 2 = 3.5. Plugging x = 3.5 back into the function, we get f(3.5) = (3.5)² - 7(3.5) + 12 = 12.25 - 24.5 + 12 = -0.25. So, the vertex is at the point (3.5, -0.25). With the roots at x = 3 and x = 4, the vertex at (3.5, -0.25), and the upward-opening nature of the parabola, we can now sketch the graph. It's a U-shaped curve intersecting the x-axis at x = 3 and x = 4, with its lowest point just below the x-axis at the vertex. This gives us a clear visual representation of the function and its key features.
d) f(x) = 5x² - 20x + 15
Now, let's tackle f(x) = 5x² - 20x + 15. Our goal remains the same determine the roots and represent the parabola. Before we dive into the quadratic formula or factoring, let's see if we can simplify this equation a bit. Notice that all the coefficients are divisible by 5! We can factor out a 5 from the entire expression: 5(x² - 4x + 3) = 0. Now, we can focus on factoring the quadratic inside the parentheses: x² - 4x + 3. We need two numbers that multiply to 3 and add up to -4. Those numbers are -1 and -3! (-1) * (-3) = 3 and (-1) + (-3) = -4. So, we can factor the quadratic as (x - 1)(x - 3) = 0. Including the 5 we factored out earlier, our equation is now 5(x - 1)(x - 3) = 0. Setting each factor to zero (note that 5 cannot be zero), we get x - 1 = 0 or x - 3 = 0, which gives us the roots x = 1 and x = 3. Fantastic! We've found the roots with a bit of simplification. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (5), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate. Looking at the original equation, -b/2a = -(-20) / (2 * 5) = 20 / 10 = 2. Plugging x = 2 back into the function, we get f(2) = 5(2)² - 20(2) + 15 = 20 - 40 + 15 = -5. So, the vertex is at the point (2, -5). With the roots at x = 1 and x = 3, the vertex at (2, -5), and the upward-opening nature of the parabola, we can sketch the graph. It's a U-shaped curve intersecting the x-axis at x = 1 and x = 3, with its lowest point well below the x-axis at the vertex. This simplified approach made finding the roots and visualizing the parabola much easier.
e) f(x) = -2x² + 6x - 4
Let's delve into the function f(x) = -2x² + 6x - 4. Our repeated objective is to determine the roots and then graph the parabola. Similar to our previous example, let's see if we can simplify this equation before we jump into factoring or the quadratic formula. We can notice that all the coefficients are divisible by -2! Factoring out -2, we get -2(x² - 3x + 2) = 0. Now, we can focus on factoring the quadratic inside the parentheses: x² - 3x + 2. We need two numbers that multiply to 2 and add up to -3. Those numbers are -1 and -2! (-1) * (-2) = 2 and (-1) + (-2) = -3. So, we can factor the quadratic as (x - 1)(x - 2) = 0. Including the -2 we factored out earlier, our equation is now -2(x - 1)(x - 2) = 0. Setting each factor to zero (remember, -2 cannot be zero), we get x - 1 = 0 or x - 2 = 0, which gives us the roots x = 1 and x = 2. Great job! We've easily found the roots by simplifying first. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is negative (-2), the parabola opens downwards. To find the vertex, we use the formula -b/2a for the x-coordinate. Looking at the original equation, -b/2a = -6 / (2 * -2) = -6 / -4 = 1.5. Plugging x = 1.5 back into the function, we get f(1.5) = -2(1.5)² + 6(1.5) - 4 = -2(2.25) + 9 - 4 = -4.5 + 9 - 4 = 0.5. So, the vertex is at the point (1.5, 0.5). With the roots at x = 1 and x = 2, the vertex at (1.5, 0.5), and the downward-opening nature of the parabola, we can sketch the graph. It's a downward-facing curve intersecting the x-axis at x = 1 and x = 2, with its highest point at the vertex. Simplifying the equation first made the process much smoother and clearer.
f) f(x) = 4x² - 4x - 3
Let's move on to the function f(x) = 4x² - 4x - 3. As always, our goal is to determine the roots and represent the parabola. This quadratic doesn't seem to have a common factor we can pull out, so let's jump straight into considering our factoring options. We need to find two numbers that multiply to (4 * -3) = -12 and add up to -4. This might take a bit more thought! The factors of -12 that come to mind are -6 and 2, since (-6) * (2) = -12 and (-6) + (2) = -4. Perfect! Now, we'll use these numbers to rewrite the middle term and factor by grouping. We can rewrite -4x as -6x + 2x, so our equation becomes 4x² - 6x + 2x - 3 = 0. Now, let's factor by grouping. From the first two terms, we can factor out 2x, giving us 2x(2x - 3). From the last two terms, we can factor out 1, giving us 1(2x - 3). Now, we have 2x(2x - 3) + 1(2x - 3) = 0. Notice the common factor of (2x - 3)! We can factor that out, giving us (2x - 3)(2x + 1) = 0. Setting each factor to zero, we get 2x - 3 = 0 or 2x + 1 = 0. Solving for x, we get x = 3/2 and x = -1/2. Excellent! We've found the roots through factoring by grouping. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (4), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate. In our case, -b/2a = -(-4) / (2 * 4) = 4 / 8 = 1/2. Plugging x = 1/2 back into the function, we get f(1/2) = 4(1/2)² - 4(1/2) - 3 = 4(1/4) - 2 - 3 = 1 - 2 - 3 = -4. So, the vertex is at the point (1/2, -4). With the roots at x = -1/2 and x = 3/2, the vertex at (1/2, -4), and the upward-opening nature of the parabola, we can sketch the graph. It's a U-shaped curve intersecting the x-axis at our calculated roots, with its lowest point well below the x-axis at the vertex. This example highlights the power of factoring by grouping when direct factoring isn't immediately obvious.
g) f(x) = 9x² - 16
Let's tackle the function f(x) = 9x² - 16. Our mission remains the same determine the roots and represent the parabola. This one looks a bit different from the others, doesn't it? We're missing the 'x' term! This is a special case called a difference of squares. We can recognize that 9x² is (3x)² and 16 is 4². The difference of squares pattern tells us that a² - b² = (a + b)(a - b). Applying this to our function, we can factor it as (3x + 4)(3x - 4) = 0. Setting each factor to zero, we get 3x + 4 = 0 or 3x - 4 = 0. Solving for x, we get x = -4/3 and x = 4/3. Fantastic! We've quickly found the roots using the difference of squares pattern. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (9), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate. But wait! What's 'b' in this case? Since there's no 'x' term, b = 0. So, the x-coordinate of the vertex is -0 / (2 * 9) = 0. Plugging x = 0 back into the function, we get f(0) = 9(0)² - 16 = -16. So, the vertex is at the point (0, -16). With the roots at x = -4/3 and x = 4/3, the vertex way down at (0, -16), and the upward-opening nature of the parabola, we can sketch the graph. It's a U-shaped curve intersecting the x-axis at our calculated roots, with its lowest point far below the x-axis at the vertex. This example beautifully demonstrates how recognizing special patterns like the difference of squares can simplify the process of finding roots.
h) f(x) = x² + 2x - 8
Let's move on to the function f(x) = x² + 2x - 8. We're on a roll with our mission to determine the roots and represent the parabola. Let's try factoring this quadratic. We need to find two numbers that multiply to -8 and add up to 2. Thinking about the factors of -8, we quickly realize that 4 and -2 fit the bill! (4) * (-2) = -8 and (4) + (-2) = 2. So, we can factor the quadratic as (x + 4)(x - 2) = 0. Setting each factor to zero, we get x + 4 = 0 or x - 2 = 0, which gives us the roots x = -4 and x = 2. Excellent! We've found the roots with straightforward factoring. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (1), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate, which is -2 / (2 * 1) = -1. Plugging x = -1 back into the function, we get f(-1) = (-1)² + 2(-1) - 8 = 1 - 2 - 8 = -9. So, the vertex is at the point (-1, -9). With the roots at x = -4 and x = 2, the vertex at (-1, -9), and the upward-opening nature of the parabola, we can now sketch the graph. It's a U-shaped curve intersecting the x-axis at x = -4 and x = 2, with its lowest point well below the x-axis at the vertex. This example reinforces the ease and efficiency of factoring when the quadratic is set up nicely for it.
i) f(x) = 2x² + x - 10
Now, let's dive into the function f(x) = 2x² + x - 10. Our objective, as always, is to determine the roots and then sketch the parabola. This quadratic doesn't have a common factor we can easily pull out, and the numbers are a bit bigger, so let's think carefully about factoring. We need to find two numbers that multiply to (2 * -10) = -20 and add up to 1 (the coefficient of the x term). After a bit of thought, we realize that 5 and -4 work perfectly! (5) * (-4) = -20 and (5) + (-4) = 1. Now, let's rewrite the middle term using these numbers and factor by grouping. We can rewrite x as 5x - 4x, so our equation becomes 2x² + 5x - 4x - 10 = 0. Now, let's factor by grouping. From the first two terms, we can factor out x, giving us x(2x + 5). From the last two terms, we can factor out -2, giving us -2(2x + 5). Now, we have x(2x + 5) - 2(2x + 5) = 0. Notice the common factor of (2x + 5)! We can factor that out, giving us (2x + 5)(x - 2) = 0. Setting each factor to zero, we get 2x + 5 = 0 or x - 2 = 0. Solving for x, we get x = -5/2 and x = 2. Fantastic! We've found the roots through factoring by grouping. Now, let's gather the information we need to sketch the parabola. Since the coefficient of the x² term is positive (2), the parabola opens upwards. To find the vertex, we use the formula -b/2a for the x-coordinate. In our case, -b/2a = -1 / (2 * 2) = -1/4. Plugging x = -1/4 back into the function, we get f(-1/4) = 2(-1/4)² + (-1/4) - 10 = 2(1/16) - 1/4 - 10 = 1/8 - 1/4 - 10 = -81/8. So, the vertex is at the point (-1/4, -81/8). With the roots at x = -5/2 and x = 2, the vertex way down at (-1/4, -81/8), and the upward-opening nature of the parabola, we can sketch the graph. It's a U-shaped curve intersecting the x-axis at our calculated roots, with its lowest point far below the x-axis at the vertex. This example further demonstrates the usefulness of factoring by grouping when dealing with more complex quadratics.
j) f(x) = -3x² + 5x - 6
Finally, let's tackle the function f(x) = -3x² + 5x - 6. Our ultimate goal remains the same determine the roots and graph the parabola. This quadratic doesn't have a common factor we can easily pull out, and the numbers don't immediately suggest a simple factoring. So, let's bring out the quadratic formula! For an equation of the form ax² + bx + c = 0, the roots are given by x = (-b ± √(b² - 4ac)) / 2a. In our case, a = -3, b = 5, and c = -6. Plugging these values into the formula, we get x = (-5 ± √(5² - 4 * -3 * -6)) / (2 * -3) = (-5 ± √(25 - 72)) / -6 = (-5 ± √(-47)) / -6. Wait a minute! We have a negative number under the square root! This means the discriminant (b² - 4ac) is negative, and the roots are complex numbers. Complex roots mean that the parabola does not intersect the x-axis. So, this quadratic function has no real roots. Now, let's gather the information we need to sketch the parabola, even though we don't have real roots. Since the coefficient of the x² term is negative (-3), the parabola opens downwards. To find the vertex, we use the formula -b/2a for the x-coordinate. In our case, -b/2a = -5 / (2 * -3) = -5 / -6 = 5/6. Plugging x = 5/6 back into the function, we get f(5/6) = -3(5/6)² + 5(5/6) - 6 = -3(25/36) + 25/6 - 6 = -25/12 + 50/12 - 72/12 = -47/12. So, the vertex is at the point (5/6, -47/12). With no real roots, the vertex at (5/6, -47/12), and the downward-opening nature of the parabola, we can sketch the graph. It's a downward-facing curve that does not intersect the x-axis, with its highest point at the vertex. This example is a great reminder that not all quadratic functions have real roots, and the quadratic formula helps us identify these cases.
Wrapping Up
Phew! We've tackled a whole bunch of quadratic functions, found their roots (or lack thereof!), and sketched their parabolas. We used factoring, the quadratic formula, and the difference of squares pattern to find the roots, and we used the coefficient of the x² term and the vertex to determine the shape and position of the parabolas. Remember, understanding the roots and the vertex is key to representing these functions graphically. Keep practicing, and you'll become a parabola pro in no time! Remember to use bold text, italic text, and strong text to emphasize the important keywords and key concepts when you write. You guys did great! Keep up the awesome work!
