Question

For x=4 the functiom f(x) =x2+bx+c has the minimum value of_9.find c

Answer 1

Answer:

$c = 25$

Step-by-step explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^{2} + bx + c$

It's vertex is the point $(x_{v}, y_{v})$

In which

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

Where

$\Delta = b^2-4ac$

If a>0, the vertex is a minumum point, that is, the minimum value happens at $x_{v}$, and it's value is $y_{v}$.

In this question:

$a = 1$

Minimum value at $x = 4$ means that $x_v = 4$. So

$x_{v} = -\frac{b}{2a}$

$4 = -\frac{b}{2}$

$b = -8$

So

$f(x) = x^2 - 8x + c$

Minimum value of 9:

$y_v = 9$

So

$y_{v} = -\frac{\Delta}{4a}$

$9 = -\frac{\Delta}{4}$

$-\Delta = 36$

$\Delta = -36$

$b^2 - 4ac = -36$

$(-8)^2 - 4c = -36$

$-4c = -100$

$4c = 100$

$c = \frac{100}{4}$

$c = 25$