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

Accepted Solution

[tex]y=2(x-7)^{2}-4[/tex] is vertex form of equation [tex]\frac{1}{2}(y+4)=(x-7)^{2}[/tex]Solution: Need to determine vertex form of the following equation  [tex]\frac{1}{2}(y+4)=(x-7)^{2}[/tex]Generic Vertex form of a quadratic equitation is as follows [tex]\mathrm{y}=\mathrm{a}(x-\mathrm{h})_{n}^{2}+\mathrm{k} \quad \text { where }(\mathrm{h}, \mathrm{k}) \text { are vertex }[/tex]So what we have to do is first make coefficient of y = 1 in our equation. [tex]y+4=2(x-7)^{2}[/tex]Now on Left hand side keep only y and move all the remaining term to right hand side [tex]\Rightarrow y=2 (x-7)^2-4\Rightarrow(1)[/tex]On comparing equation (1) with generic vertex form equation we can say that    In our case a = 2, h = 7 and k = -4 Hence [tex]y=2(x-7)^{2}-4[/tex] is vertex form of equation [tex]\frac{1}{2}(y+4)=(x-7)^{2}.[/tex]