# How do you find the vertex and intercepts for y = x^2 - 10x +17?

Aug 18, 2017

vertex: $\left(5 , - 8\right)$

x-intercepts: $\left(5 + 2 \sqrt{2} , 0\right) ,$$\left(5 - 2 \sqrt{2} , 0\right)$

y-intercept: $\left(0 , 17\right)$

Refer to the explanation for the process.

#### Explanation:

Given:

$y = {x}^{2} - 10 x + 17$ is a quadratic equation in standard form:

$a {x}^{2} - 10 x + 17$,

where:

$a = 1$, $b = - 10$, and $c = 17$

Vertex: the maximum or minimum point of the parabola

axis of symmetry and $x$ value for vertex: $x = \frac{- b}{2 a}$

$x = \frac{- \left(- 10\right)}{2 \cdot 1}$

$x = \frac{10}{2}$

$x = 5$

To find the $y$ value of the vertex, substitute $5$ for $x$ into the equation and solve for $y$.

$y = 1 \times {5}^{2} - 10 \left(5\right) + 17$

$y = 25 - 50 + 17$

$y = - 8$

The vertex is $\left(5 , - 8\right)$. Since $a > 0$, the vertex is the minimum point of the parabola, and the parabola opens upward.

X-Intercepts: the value of $x$ when $y = 0$

To find the x-intercepts, substitute $0$ for $y$ and solve for $x$ using the quadratic formula.

$0 = {x}^{2} - 10 x + 17$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in the known values.

$x = \frac{- \left(- 10\right) \pm \sqrt{- {10}^{2} - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$

Simplify.

$x = \frac{10 \pm \sqrt{100 - 68}}{2}$

$x = \frac{10 \pm \sqrt{32}}{2}$

Prime factorize $32$.

$x = \frac{10 \pm \sqrt{\textcolor{red}{2} \times \textcolor{red}{2} \times \textcolor{b l u e}{2} \times \textcolor{b l u e}{2} \times 2}}{2}$

Simplify.

$x = \frac{10 \pm \textcolor{red}{2} \times \textcolor{b l u e}{2} \sqrt{2}}{2}$

$x = \frac{10 \pm 4 \sqrt{2}}{2}$

Simplify.

$x = \frac{{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}}}^{5} \pm {\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}^{2} \sqrt{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} ^ 1$

$x = 5 \pm 2 \sqrt{2}$

$x = 5 + 2 \sqrt{2} ,$$5 - 2 \sqrt{2}$

x-intercepts: $\left(5 + 2 \sqrt{2} , 0\right) ,$$\left(5 - 2 \sqrt{2} , 0\right)$

Y-Intercept: the value of $y$ when $x = 0$

To find the y-intercept, substitute $0$ for $x$ and solve for $y$.

$y = 1 \times {0}^{2} - 10 \left(0\right) + 17$

$y = 0 + 0 + 17$

$y = 17$

y-intercept is $\left(0 , 17\right)$

Summary

axis of symmetry: $5$

vertex: $\left(5 , - 8\right)$

x-intercepts: $\left(5 + 2 \sqrt{2} , 0\right) ,$$\left(5 - 2 \sqrt{2} , 0\right)$

y-intercept: $\left(0 , 17\right)$

graph{y=x^2-10x+17 [-15.68, 16.34, 5.38, 21.4]}