Chapter 7.5, Problem 9WE

To determine

Draw the graph .Find the vertex , axis of symmetry and all the intercepts.

Answer to Problem 9WE

Vertex : $(-1,3)$

Axis of Symmetry : x = -1

y -intercept : (0,5)

Explanation of Solution

Given Information:

Given equation is :

  $y-3=2{\left(x+1\right)}^2$

Concept Used :

• Parent Function:

  $y=x^2$

• Vertex = (0,0)
• Type of Graph:

Upward Parabola

• Axis of Symmetry: Passes through the vertex and divides the parabola into two congruent halves.

  $x=0$

• y -intercept = The y coordinate of the point at which the graph intersects y -axis , x = 0.
• x -intercept = The x coordinate of the point at which the graph intersects x -axis , y = 0 .

Calculation:

• In the equation $y-3=2{\left(x+1\right)}^2$ ; comparing it with standard form of parabola equation $Y=X^2$, opens upward , has vertex (0,0) , axis of symmetry is X = 0.
• $y-3=Y;2{\left(x+1\right)}^2={\left(\sqrt{2}\left(x+1\right)\right)}^2=X^2$
• Vertex :

  $\begin{array}{l}(X,Y)=(0,0)\\ \Rightarrow y-3=Y=0;\sqrt{2}\left(x+1\right)=X=0\\ \Rightarrow y=3;x=-1\\ \Rightarrow Vertex\equiv (-1,3)\end{array}$

• Find Axis of Symmetry :

  $\begin{array}{l}X=\mathrm{0........................}{\text{[axis of symmetry of Y=X}}^2]\\ \Rightarrow \sqrt{2}\left(x+1\right)=\mathrm{0........}[X=\sqrt{2}\left(x+1\right)]\\ \Rightarrow x=-\mathrm{1..................}[divide\text{ each side by }\sqrt{2}\text{ then subtracrt 1 from each side]}\end{array}$

The equation of axis of symmetry is x = -1

• Find y -intercept :

The y -coordinate of the point at which the graph cuts the y -axis is the y -intercept.

Hence , substituting x =0 in the equation $y-3=2{\left(x+1\right)}^2$ , we will get the y -intercept.

  $\begin{array}{l}y-3=2{\left(x+1\right)}^2\\ \Rightarrow y-3=2{\left(0+1\right)}^2\mathrm{.............}[x=0]\\ \Rightarrow y-3=2{\left(1\right)}^2\\ \Rightarrow y=2+\mathrm{3.........................}[\text{add 3 each side]}\\ \Rightarrow y=5\end{array}$

So, y -intercept = (0,5)

• Find x -intercept :

The x -coordinate of the point at which the graph cuts the x -axis is the x -intercept.

Hence , substituting y =0 in the equation $y-3=2{\left(x+1\right)}^2$ , we will get the x -intercept.

  $\begin{array}{l}y-3=2{\left(x+1\right)}^2\\ \Rightarrow 0-3=2{\left(x+1\right)}^2\mathrm{.............}[y=0]\\ \Rightarrow -3=2{\left(x+1\right)}^2\end{array}$

Which is not possible since square of a number is always non-negative.

So, x -intercept does not exist.

Use the above information to graph the equation $y-3=2{\left(x+1\right)}^2$: