What is the vertex of 7y= 12(x-15)^2 +12?

The vertex happens to be
$\left(x , y\right) = \left(15 , \frac{12}{7}\right)$

Explanation:

The given equation is:
$7 y = 12 {\left(x - 15\right)}^{2} + 12$
The curve is symmetrical about the x axis
Differentiating the equation wrt x
$7 \frac{\mathrm{dy}}{\mathrm{dx}} = 12 \left(2\right) \left(x - 15\right) + 0$
The vertex coresponds to the point where the slope is zero.
Equating $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$7 \left(0\right) = 24 \left(x - 15\right)$
ie
$24 \left(x - 15\right) = 0$
$x - 15 = 0$
$x = 15$
Substituting for x in the equation of the curve
$7 y = 12 \left(15 - 15\right) + 12$
$7 y = 12$
$y = \frac{12}{7}$
Thus, the vertex happens to be
$\left(x , y\right) = \left(15 , \frac{12}{7}\right)$

Feb 9, 2018

$\text{vertex } = \left(15 , \frac{12}{7}\right)$

Explanation:

$\text{divide both sides by 7}$

$\Rightarrow y = \frac{12}{7} {\left(x - 15\right)}^{2} + \frac{12}{7}$

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$y = \frac{12}{7} {\left(x - 15\right)}^{2} + \frac{12}{7} \text{ is in vertex form}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(15 , \frac{12}{7}\right)$