Start with writing the function in standard form. Then find its vertex to graph it.
Inverse: f^(- 1)(x)=± sqrt(x)-1/2 Graph:
Practice makes perfect
Before we can find the inverse of the given function, we need to replace f(x) with y.
f(x)=(2x+1)^2 ⇔ y=(2x+1)^2
To algebraically determine the inverse of the given equation, we exchange x and y and solve for y.
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Given Equation & Inverse Equation [0.3em]
y=(2 x+1)^2 & x=(2 y+1)^2
The result of isolating y in the new equation will be the inverse of the given function.
Now we have the inverse of the given function.
f^(- 1)(x)=± sqrt(x)-1/2
Graphing the Function
Because the given function is a parabola, to graph it we should first determine its vertex. However, the function is in neither standard form nor vertex form. Let's first rewrite it in standard form.
Now that we have the function in standard form, let's examine the coefficients of the terms to find its vertex and graph it.
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Standard Form & Function
y= ax^2+ bx+ c & y= 4x^2+ 4x+ 1
In this form, if a is positive, the parabola opens upward. If a is negative, the parabola opens downward. Since 4>0, the parabola of this function opens upward. To find the vertex, we first need to find the x-coordinate of the vertex.
x=-b/2 a
Let's find it!
Thus, the vertex of the parabola is ( 12,0). To graph the parabola let's choose two more points, one on either side of the vertex. Let's use x=-1 and x=0. By substituting these coordinates into the function, we can find the y-coordinates.
x
4x^2-12x+8
y
Point
-1
4( -1)^2+4( -1)+1
1
(-1,1)
- 1/2
4( - 1/2)^2+4( - 1/2)+1
0
(-1/2,0)
0
4( 0)^2+4( 0)+1
1
(0,1)
Let's plot the points and connect them to graph the parabola.
Graphing the Inverse of the Function
Finally, we can graph the inverse of the function by reflecting the parabola across y=x. This means that we should interchange the x- and y- coordinates of the points that are on the parabola.
