Question #N715
The function $f(x) = ax^2 + bx + c$, where $a$, $b$, and $c$ are constants, has a minimum value of 5 at $x = 2$. If the graph of $f(x)$ intersects the y-axis at $(0, 9)$, what is the value of $c$?
A. 1
B. 5
C. 9
D. 13
Correct Answer is: C
Since the function has a minimum value of 5 at x = 2, the vertex of the parabola is at (2, 5). We can use the vertex form of a quadratic equation to write the equation: $f(x) = a(x - 2)^2 + 5$. We know the graph intersects the y-axis at (0, 9), which means f(0) = 9. Substituting these values into the equation, we get: 9 = a(0 - 2)^2 + 5. Simplifying the equation, we get: 9 = 4a + 5. Solving for a, we get a = 1. Substituting this value of a back into the vertex form, we get: $f(x) = (x - 2)^2 + 5$. Expanding the equation, we get: $f(x) = x^2 - 4x + 4 + 5$, or $f(x) = x^2 - 4x + 9$. Therefore, the value of c is 9.