Question

1/2 (y+4)=(x-7)^2 in vertex form step by step

Answer 1

$y=2(x-7)^{2}-4$ is vertex form of equation $\frac{1}{2}(y+4)=(x-7)^{2}$

Solution:

Need to determine vertex form of the following equation  

$\frac{1}{2}(y+4)=(x-7)^{2}$

Generic Vertex form of a quadratic equitation is as follows

$\mathrm{y}=\mathrm{a}(x-\mathrm{h})_{n}^{2}+\mathrm{k} \quad \text { where }(\mathrm{h}, \mathrm{k}) \text { are vertex }$

So what we have to do is first make coefficient of y = 1 in our equation.

$y+4=2(x-7)^{2}$

Now on Left hand side keep only y and move all the remaining term to right hand side

$\Rightarrow y=2 (x-7)^2-4\Rightarrow(1)$

On comparing equation (1) with generic vertex form equation we can say that    In our case a = 2, h = 7 and k = -4

Hence $y=2(x-7)^{2}-4$ is vertex form of equation $\frac{1}{2}(y+4)=(x-7)^{2}.$