Finding the Vertex of a Quadratic Function

Oct 13, 2025 by sorusgentrification.com 112 views

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Introduction

In mathematics, a quadratic function is a polynomial function of degree two, which means the highest power of the variable is two. The general form of a quadratic function is $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants. The vertex of a quadratic function is the maximum or minimum point of the parabola it represents. In this article, we will discuss how to find the vertex of a quadratic function, with a focus on the function $f(x) = (x-8)(x-2)$ .

What is the Vertex?

The vertex of a quadratic function is the point at which the function changes from decreasing to increasing or vice versa. It is the maximum or minimum point of the parabola, and it is represented by the coordinates $(h, k)$ , where $h$ is the x-coordinate and $k$ is the y-coordinate.

Finding the Vertex of a Quadratic Function

To find the vertex of a quadratic function, we can use the formula $(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$ , where $a$ , $b$ , and $c$ are the coefficients of the quadratic function. However, this formula is only applicable when the quadratic function is in the form $f(x) = ax^2 + bx + c$ . If the quadratic function is in the form $f(x) = (x-r_1)(x-r_2)$ , where $r_1$ and $r_2$ are the roots of the function, we can find the vertex by using the formula $(h, k) = (\frac{r_1+r_2}{2}, f(\frac{r_1+r_2}{2}))$ .

Finding the Vertex of the Function $f(x) = (x-8)(x-2)$

To find the vertex of the function $f(x) = (x-8)(x-2)$ , we need to first expand the function to get it in the form $f(x) = ax^2 + bx + c$ . Expanding the function, we get:

$f(x) = (x-8)(x-2)$ $f(x) = x^2 - 2x - 8x + 16$ $f(x) = x^2 - 10x + 16$

Now that we have the function in the form $f(x) = ax^2 + bx + c$ , we can find the vertex by using the formula $(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$ . In this case, $a = 1$ and $b = -10$ , so we can plug these values into the formula to get:

$h = -\frac{-10}{2(1)}$ $h = 5$

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by plugging the x-coordinate into the function:

$k = f(5)$ $k = (5)^2 - 10(5) + 16$ $k = 25 - 50 + 16$ $k = -9$

Therefore, the vertex of the function $f(x) = (x-8)(x-2)$ is $(5, -9)$ .

Conclusion

In conclusion, finding the vertex of a quadratic function is an important concept in mathematics. The vertex is the maximum or minimum point of the parabola, and it is represented by the coordinates $(h, k)$ . We can find the vertex of a quadratic function by using the formula $(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$ or by using the formula $(h, k) = (\frac{r_1+r_2}{2}, f(\frac{r_1+r_2}{2}))$ if the function is in the form $f(x) = (x-r_1)(x-r_2)$ . In this article, we discussed how to find the vertex of the function $f(x) = (x-8)(x-2)$ , and we found that the vertex is $(5, -9)$ .

Example Problems

Problem 1

Find the vertex of the function $f(x) = (x-3)(x-5)$ .

Solution

To find the vertex of the function $f(x) = (x-3)(x-5)$ , we need to first expand the function to get it in the form $f(x) = ax^2 + bx + c$ . Expanding the function, we get:

$f(x) = (x-3)(x-5)$ $f(x) = x^2 - 5x - 3x + 15$ $f(x) = x^2 - 8x + 15$

Now that we have the function in the form $f(x) = ax^2 + bx + c$ , we can find the vertex by using the formula $(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$ . In this case, $a = 1$ and $b = -8$ , so we can plug these values into the formula to get:

$h = -\frac{-8}{2(1)}$ $h = 4$

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by plugging the x-coordinate into the function:

$k = f(4)$ $k = (4)^2 - 8(4) + 15$ $k = 16 - 32 + 15$ $k = -1$

Therefore, the vertex of the function $f(x) = (x-3)(x-5)$ is $(4, -1)$ .

Problem 2

Find the vertex of the function $f(x) = x^2 + 6x + 8$ .

Solution

To find the vertex of the function $f(x) = x^2 + 6x + 8$ , we can use the formula $(h, k) = (-\frac{b}{2a}, f(-\frac{b}{2a}))$ . In this case, $a = 1$ and $b = 6$ , so we can plug these values into the formula to get:

$h = -\frac{6}{2(1)}$ $h = -3$

Now that we have the x-coordinate of the vertex, we can find the y-coordinate by plugging the x-coordinate into the function:

$k = f(-3)$ $k = (-3)^2 + 6(-3) + 8$ $k = 9 - 18 + 8$ $k = -1$

Therefore, the vertex of the function $f(x) = x^2 + 6x + 8$ is $(-3, -1)$ .