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Question: How do you graph \(y={{\left( x+3 \right)}^{2}}-1\) ?...

How do you graph y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1 ?

Explanation

Solution

We are given y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1 . First we know that the equation is a quadratic equation as it has square over ‘x’ only, so it is parabola, to sketch the graph of this we will need following this –
We find vertex, y-intercept, x-intercept.
We will then locate these points on the graph, once they are located, we will join the point and sketch the required graph.
We use the general equation y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k to compare so that we find (h,k)\left( h,k \right) .

Complete step by step answer:
We are given an equation as y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1.
We are asked to draw the graph of this equation.
As we can see that it is a quadratic equation in which ‘x’ has the highest power of 2 while ‘y’ is only linear.
So, our equation is a kind of parabola.
As we have to plot this we should know this to plot any parabola, we should know about the vertex of the parabola, axis of symmetry of the parabola and also know about some of the points that lie on the particular parabola.
In general, parabola in vertex term is given as –
(y−k)=a(x−h)2\left( y-k \right)=a{{\left( x-h \right)}^{2}} or y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k
Where (h,k)\left( h,k \right) are the coordinates of the vertex of the parabola.
Now, to sketch the parabola y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k
We follow following steps:

  1. First we find vertex of the parabola
  2. Find the y intercept by putting x=0 and solve for the value of ‘y’.
  3. Find the ‘x’ intercept by putting y=0 and solve the equation for ‘x’.
  4. Then we plot vertex, ‘x’ intercept, ‘y’ intercept and join the points to form the graph.
    Now, we are given that y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1
    So, we compare it with y=a(x−h)2+ky=a{{\left( x-h \right)}^{2}}+k
    We have –
    h=−3h=-3 and k=−1k=-1 .
    Now, as vertex is given as (h,k)\left( h,k \right) so we get our vertex are (−3,−1)\left( -3,-1 \right)
    Secondly we find the value of ‘y’ intercept.
    We put x=0 in our equation, y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1 and solve for ‘y’.
    We get –
    y=(0+3)2−1y={{\left( 0+3 \right)}^{2}}-1
    As 0+3=30+3=3 so –
    y=32−1y={{3}^{2}}-1
    =9−1=9-1
    y=8y=8
    So, we get –
    y=8y=8
    So, ‘y’ intercept is 8
    Graph of ‘y’ cut y-axis at 8
    Now we look for ‘x’ intercept
    We put y=0, in y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1
    We get –
    0=(x+3)2−10={{\left( x+3 \right)}^{2}}-1
    We solve for ‘x’.
    So,
    1=(x+3)21={{\left( x+3 \right)}^{2}}
    ±1=x+3 x=±1−3 \begin{aligned} & \pm \sqrt{1}=x+3 \\\ & x=\pm \sqrt{1}-3 \\\ \end{aligned}
    By simplifying, we get –
    x=±1−3x=\pm 1-3
    We take +1, then we get x=+1−3=−2x=+1-3=-2 and when we take -1, we get x=−1−3=−4x=-1-3=-4
    So our x intercepts are -2 and -4.
    So,
    Graph of x-axis at -2 and -4.
    Now, as our equation has a square over ‘x’ so the axis of symmetry is alone y-axis.
    So, using vertex as (h,k)=(−3,−1)\left( h,k \right)=\left( -3,-1 \right)
    Y intercept (0,8)\left( 0,8 \right)
    X-intercept (−2,0)\left( -2,0 \right) and (−4,0)\left( -4,0 \right)
    We will put the point on a graph and sketch our required graph.

Note:
Remember that if we pick the wrong axis of symmetry then we will get the incorrect graph also that we need to pick the correct value of h and k we have –
y=(x+3)2−1y={{\left( x+3 \right)}^{2}}-1 than we have y=[x−(−3)]2+(−1)y={{\left[ x-\left( -3 \right) \right]}^{2}}+\left( -1 \right)
So here we compare with y=(x−h)2+ky={{\left( x-h \right)}^{2}}+k
We get –
h=−3h=-3 and 5=75=7 , we need to precise if we recklessly pick h=3h=3 and k=1k=1 it will give an incorrect graph.