# How do you find the vertex and intercepts for y = x^2 - 8x + 18?

Nov 23, 2015

Vertex={(4,2)}, intercept={(0,18)}

#### Explanation:

$y = {x}^{2} - 8 x + 18$
$\implies y ' = 2 x - 8$
For $y ' = 0$,
$2 x - 8 = 0$
$\implies x = 4$
$\implies y = {4}^{2} - 8 \cdot 4 + 18$
$= 16 - 32 + 18$
$= 2$
$\implies v e r t e x = \left\{\left(4 , 2\right)\right\}$
$\Delta = {\left(- 8\right)}^{2} - 4 \cdot 1 \cdot 18$
$= - 8$
$\implies \Delta < 0$
For $y = 0$,
${x}^{2} - 8 x + 18 = 0$
$\implies$ $x$ is imaginary
For $x = 0$,
$y = 18$
$\implies$ intercept={(0,18)}

Nov 23, 2015

vertex: $\left(4 , 2\right)$
y-intercept: $18$
there is no x-intercept

#### Explanation:

Converting the given equation into vertex form: $y = m {\left(x - a\right)}^{2} + b$ with vertex at $\left(a , b\right)$

$y = {x}^{2} - 8 x + 18$

$\rightarrow y = {x}^{2} - 8 x + 16 + 2$

$\rightarrow y = 1 {\left(x - 4\right)}^{2} + 2$
$\textcolor{w h i t e}{\text{XXX}} w i t h v e r t e x a t$(4,2)

y-intercept is the value of $y$ when $x = 0$
$\Rightarrow$ y-intercept = 18

x-intercept is the value of $x$ when $y = 0$
i.e. when ${x}^{2} - 8 x + 18 = 0$
but checking the discriminant (${b}^{2} - 4 a c$ using the standard form)
we see that there are no solutions for $x$ since the discriminant is $< 0$