Question

The graph of a parabola is given. Match graph to its equation.

$$\begin{gathered} A)y^2=4xC)y^2=-4xE){(y-1)}^2=4(x-1)G){(y-1)}^2=-4(x-1) \\ B)x^2=4yD)x^2=-4yF){(x+1)}^2=4(y+1)H){(x+1)}^2=4(y+1) \end{gathered}$$

Short answer

The correct match is (E).

Step-by-step solution

Step 1. Given information.

We have given graph is,

and given equations are$$\begin{gathered} A)y^2=4xC)y^2=-4xE){(y-1)}^2=4(x-1)G){(y-1)}^2=-4(x-1) \\ B)x^2=4yD)x^2=-4yF){(x+1)}^2=4(y+1)H){(x+1)}^2=4(y+1) \end{gathered}$$

Step 2. Concept used.

General equation of the parabola is,${\left(x-h\right)}^2=4a\left(y-k\right)\hspace{0.17em}and\hspace{0.17em}{\left(y-k\right)}^2=4a\left(x-h\right)$

Step 3. Explanation.

We have given graph is,

and vertex of the parabola is (1, 1).

this graph is the parabola opens to right.

Using equation of the parabola and vertex,

$$\begin{gathered} {\left(y-k\right)}^2=4a\left(x-h\right) \\ \Rightarrow {\left(y-1\right)}^2=4a\left(x-1\right) \end{gathered}$$

Graph is opens to the right hence so a is positive (a > 0).

Therefore equation ${\left(y-1\right)}^2=4\left(x-1\right)$

is the equation of given graph.

Step 4. Conclusion.

Hence, equation of the graph

is${\left(y-1\right)}^2=4\left(x-1\right)$.