# How do you find the maximum or minimum of y+3x^2=9?

May 23, 2018

Maximum value of $y = 9$ at $x = 0$

#### Explanation:

$y + 3 {x}^{2} = 9 \mathmr{and} y = - 3 {x}^{2} + 9 \mathmr{and} y = - 3 {\left(x - 0\right)}^{2} + 9$.

This is equation of parabola opening downward since coefficient of

${x}^{2}$ is negative. So minimum value will be $- \infty$ and

maximum value will be at vertex. Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = 0 , k = 9 \therefore$ Vertex is at $\left(0 , 9\right)$

Therefore the minimum value of function is $- \infty$ and

maximum value of function is $y = 9$ at $x = 0$

graph{y+ 3x^2=9 [-22.5, 22.5, -11.25, 11.25]} [Ans]