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

Sep 4, 2017

$\text{see explanation}$

#### Explanation:

$\text{the standard form of a parabola with the y-axis as it's}$
$\text{principal axis and opening vertically is}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{x}^{2} = 4 p y} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{this parabola has it's vertex at the origin } \left(0 , 0\right)$

• " if 4p ">0" then opens vertically up "uuu

• " if 4p "<0" then opens vertically down "nnn

${x}^{2} = 8 y \text{ is in this form}$

$\Rightarrow 4 p = 8$

$\text{since 4p ">0" then opens up}$
graph{x^2=8y [-10, 10, -5, 5]}

Sep 4, 2017

Vertex $= \left(0 , 0\right)$ which is also the only intercept of the $x y -$axes.

#### Explanation:

${x}^{2} = 8 y \to y = \frac{1}{8} {x}^{2}$

The graph of $y$ is a parabola, with a vertex at the absolute extremum of $y$

Since the coefficient of ${x}^{2} = \frac{1}{8} > 0$ the vertex of $y$ will be an absolute minimum.

Since ${x}^{2} \ge 0 \forall x \in \mathbb{R} \to {y}_{\min} = y \left(0\right)$

$\therefore {y}_{\min} = \frac{1}{8} \times 0 = 0$

Hence, the vertex of $y = \left(0 , 0\right)$

Since the vertex is the absolute minimum of $y$ there can be no other intercepts than $\left(0 , 0\right)$

This result can be seen from the graph of $y$ below.

graph{x^2/8 [-16.11, 15.93, -2.15, 13.87]}