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Graph:
x | x^2+x-6 | f(x) |
---|---|---|
-4 | ( -4)^2+( -4)-6 | 6 |
-3 | ( -3)^2+( -3)-6 | 0 |
-2 | ( -2)^2+( -2)-6 | -4 |
-1 | ( -1)^2+( -1)-6 | -6 |
-1/2 | ( -1/2)^2+( -1/2)-6 | -25/4 |
0 | ( 0)^2+ 0-6 | -6 |
1 | ( 1)^2+ 1-6 | -4 |
2 | ( 2)^2+ 2-6 | 0 |
3 | ( 3)^2+ 3-6 | 6 |
Let's plot the points and draw a parabola that passes through these points and complete the graph of the function.
Use the Zero Product Property
(I): LHS+2=RHS+2
(II): LHS-3=RHS-3
a | a(x-2)(x+3) | f(x) |
---|---|---|
4 | 4(x-2)(x+3) | 4x^2+4x-24 |
-3 | -3(x-2)(x+3) | -3x^2-3x+18 |
1/2 | 1/2(x-2)(x+3) | 1/2x^2+1/2x-3 |
To graph the functions, we will proceed in the same way as we did in Part A. Let's first find their axes of symmetry.
f(x) | x=-b/2a | Axis of Symmetry |
---|---|---|
4x^2+ 4x -24 | x=-4/2( 4) | x=-1/2 |
-3x^2 -3x+ 18 | x=--3/2( -3) | x=-1/2 |
1/2x^2+ 1/2x -3 | x=-1/2/2( 1/2) | x=-1/2 |
Now, we can graph these functions as we did in Part A. Notice that the leading coefficient of the second function is negative, so the graph of it opens downward. The other two functions open upward because of the positive leading coefficient.
f(x)=a(x-p)(x-q) When a quadratic function is in factored form as shown, we can immediately determine its solutions as p and q by the Zero Product Property.