Math

QuestionFind the quadratic function with a vertex at (5,9) that increases on (,5)(-\infty,5) and decreases on (5,)(5,\infty). Options: a. f(x)=(x+5)2+9f(x)=(x+5)^{2}+9, b. f(x)=(x5)2+9f(x)=(x-5)^{2}+9, c. f(x)=(x+5)2+9f(x)=-(x+5)^{2}+9, d. f(x)=(x5)2+9f(x)=-(x-5)^{2}+9.

Studdy Solution

STEP 1

Assumptions1. The graph of the quadratic function increases through interval A and decreases through interval B. . The vertex of the graph is at the point (5,9).
3. The quadratic function is in the form f(x)=a(xh)+kf(x)=a(x-h)^{}+k, where (h,k) is the vertex of the parabola.

STEP 2

We know that the vertex of the graph is at (5,9). This means that the value of h is5 and the value of k is9 in the equation f(x)=a(xh)2+kf(x)=a(x-h)^{2}+k.

STEP 3

Substitute the values of h and k into the equation.
f(x)=a(x5)2+9f(x)=a(x-5)^{2}+9

STEP 4

We know that the graph of the function increases through interval A and decreases through interval B. This means that the parabola opens downwards. In other words, the coefficient a is negative.

STEP 5

Substitute a negative value for a in the equation.
f(x)=a(x5)2+9f(x)=-a(x-5)^{2}+9

STEP 6

Now, we can compare this equation with the given options. The equation that matches our derived equation is option d.
So, the equation that could represent this function is f(x)=(x5)2+9f(x)=-(x-5)^{2}+9.

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