# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of y=x^2-10x+2?

Oct 10, 2015

Vertex is $\left(5 , - 23\right)$
Axis of symmetry $x = 5$
Y-intercept $y = 2$
X- intercepts are $\left(9.8 , 0\right) \mathmr{and} \left(0.2 , 0\right)$

#### Explanation:

Vertex

$3 x = \frac{- b}{2 a} = \frac{- \left(- 10\right)}{2 \times 1} = \frac{10}{2} = 5$

At $x = 5$

$y = {5}^{2} - 10 \left(5\right) + 2$
$y = 25 - 50 + 2$
$= 27 - 50$
$y = - 23$
Vertex is $\left(5 , - 23\right)$
Axis of symmetry $x = 5$

Y-intercept

At $x = 0$

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

${b}^{2} - \left(4 a c\right) = \left(- {10}^{2}\right) - \left(4 \times 1 \times 2\right)$

X-intercept

At $y = 0$

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

$x = \frac{- b \pm \sqrt{{b}^{2} - \left(4 a c\right)}}{2 a}$
$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - \left(4 \times 1 \times 2\right)}}{2 \times 1}$
$x = \frac{10 \pm \sqrt{100 - 8}}{2}$
$x = \frac{10 \pm \sqrt{92}}{2}$
$x = \frac{10 \pm 9.6}{2}$
$x = \frac{10 + 9.6}{2} = \frac{19.6}{2} = 9.8$
$x = \frac{10 - 9.6}{2} = \frac{0.4}{2} = 0.2$

X- intercepts are $\left(9.8 , 0\right) \mathmr{and} \left(0.2 , 0\right)$

graph{x^2-10x+2 [-58.5, 58.55, -29.25, 29.25]}