write an equation in standard form of the parabola that has the same shape as the graph of f(x)-7x squared, but which has a maximum of 9 at x=3.
f(x)= M-k(x-3)^2
Ok, you figure out what M is and k. Remember what k does to shape.
To determine the equation in standard form for the parabola with the given conditions, we need to consider a transformation of the graph of the function f(x) = -7x^2.
Step 1: Find the vertex form of the parabola.
To shift the vertex from the origin to (3, 9), we need to add the values (h, k) to the equation. In this case, h = 3 and k = 9. Therefore, the vertex form is f(x) = a(x - 3)^2 + 9.
Step 2: Determine the value of the coefficient 'a'.
Since the parabola is the same shape as the graph of f(x) = -7x^2, we need to find the appropriate value of 'a'. We can do this by comparing the coefficient of x^2 in the original function (-7x^2) to the vertex form equation: f(x) = a(x - 3)^2 + 9.
To find 'a', set f(x) = -7x^2 and x = 0:
-7(0)^2 = a(0 - 3)^2 + 9
0 = 9a + 9
9a = -9
a = -1
Step 3: Express the equation in standard form.
Now that we have the value of 'a', we can substitute it into the vertex form equation: f(x) = -1(x - 3)^2 + 9.
Expanding and rearranging, we have:
f(x) = -(x - 3)^2 + 9
f(x) = -(x^2 - 6x + 9) + 9
f(x) = -x^2 + 6x - 9 + 9
f(x) = -x^2 + 6x
Therefore, the equation in standard form of the parabola with the same shape as f(x) = -7x^2, but with a maximum of 9 at x = 3, is f(x) = -x^2 + 6x.