How do you find the vertex and intercepts for y=2(x+2)^2+3?

1 Answer
Jun 4, 2017

Vertex is at $\left(- 2 , 3\right)$ , y-intercept is at $0 , 11$ , No x-intercept.

Explanation:

$y = 2 {\left(x + 2\right)}^{2} + 3$. Comparing with standard vertex form of equation
y=a(x-h)^2 +k ; (h,k)being vertex , we find here $h = - 2 , k = 3$
Hence vertex is at $\left(- 2 , 3\right)$

y-intercept is obtained by putting $x = 0$ in the equation.
$\therefore y = 2 {\left(0 + 2\right)}^{2} + 3 = 11$ or at $\left(0 , 11\right)$

x-intercept is obtained by putting $y = 0$ in the equation.
$\therefore 2 {\left(x + 2\right)}^{2} + 3 = 0 \mathmr{and} 2 {\left(x + 2\right)}^{2} = - 3 \mathmr{and} {\left(x + 2\right)}^{2} = - \frac{3}{2}$ or
$\left(x + 2\right) = \pm \sqrt{- \frac{3}{2}} \mathmr{and} x = - 2 \pm \sqrt{- \frac{3}{2}}$ or
$x = - 2 \pm \sqrt{\frac{3}{2}} i \therefore x$ has complex roots.
So there is no x-intercept.

graph{2(x+2)^2+3 [-40, 40, -20, 20]} [Ans]