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Graph:
x-intercepts: (-1,0) and (3,0)
y-intercepts: (0,6)
f(x) = -2(x-1)^2+8
Next, recall that the general expression of a quadratic function in a graphing form is f(x)=a(x-h)^2+k, where a ≠0. Let's note three things we can learn from this equation.f(x) = -2(x-1)^2+8 Looking at the given function we have a = -2, h = 1, and k = 8. Let's use these values to determine the mentioned properties of a parabola.
Now, let's make a table of values using five points. We want the center point to be the vertex and the remaining points to be symmetric on either side of it. We know that the points will be symmetric if the x-coordinates are equidistant from the axis of symmetry.
x | - 2(x-1)^2 + 8 | f(x)=- 2(x-1)^2 + 8 |
---|---|---|
- 1 | - 2( -1-1)^2 + 8 | 0 |
0 | - 2( 0-1)^2 + 8 | 6 |
1 | - 2( 1-1)^2 + 8 | 8 |
2 | - 2( 2-1)^2 + 8 | 6 |
3 | - 2( 3-1)^2 + 8 | 0 |
Finally, we will graph the function by plotting the points from the table. Because the graph of a quadratic function is a parabola, we will connect them with a smooth curve.
Now, notice that among the chosen points we have two x-intercepts and a y-intercept. Let's list these points. cc x-intercepts & y-intercept (-1,0), (3,0) & (0,6)
f(x) = -2(x-1)^2+8
Looking at the given function we have h = 1, and k = 8. Therefore, point (1, 8) is the vertex.x= 100
Subtract term
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x= -15
Subtract term
Calculate power
Multiply
Add terms