Question

(a) Write $x=3y^2+6y+7$in standard form and (b) use properties of the standard form to graph the equation.

Short answer

Part (a), The standard form is $x=3{(y+1)}^2+4$

Part (b), The graph is

Step-by-step solution

Part (a) Step 1: Given information

Given the function$x=3y^2+6y+7$

Part (a) Step 2: Rewriting the function

Rewriting, we get

$$\begin{gathered} x=3y^2+6y+7 \\ x=3(y^2+2y)+7 \\ x=3(y^2+2y+1^2)+4 \\ x=3{(y+1)}^2+4 \end{gathered}$$

Part (b) Step 1: Given information

Given the standard form$x=3(y+1{)}^2+4$

Part (b) Step 2: Identify the constants a, h, k and the other parameters

Comparing, we get

$$\begin{gathered} x=3(y+1{)}^2+4 \\ a=3 \\ (h,k)=(4,-1) \end{gathered}$$

Part (b) Step 3: Calculate x intercept

Calculating, we get

$$\begin{gathered} x=3(y+1{)}^2+4 \\ x=3{(0+1)}^2+4 \\ x=7 \end{gathered}$$

Part (b) Step 4: Calculating y intercept

Calculating, we get

$$\begin{gathered} x=3(y+1{)}^2+4 \\ 3{(y+1)}^2+4=0 \\ 3{(y+1)}^2=-4 \end{gathered}$$

Square can't be negative so there is no$y$intercept

Part (b) Step 5: Sketching the graph

The graph is