Question

Is my answer correct? 10 points + brainleist!

Answer 1

Answer:

your answer is incorrect.  The correct answer is  $h=-13$  and $k=13$ .

Step-by-step explanation:

If a quadratic function is $f(x)=ax^2+bx+c$ and a>0, then minimum value of the function at point $\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)$.

The given function is

$f(x)=x^2+bx+182$

Here, a=1, b=b and c=182. So.

$-\dfrac{b}{2a}=-\dfrac{b}{2(1)}=-\dfrac{b}{2}$

Put $x=-\dfrac{b}{2}$ in the given function to find the minimum value of the function.

$f(-\dfrac{b}{2})=(-\dfrac{b}{2})^2+b(-\dfrac{b}{2})+182$

We know that minimum value is 13. So,

$13=\dfrac{b^2}{4}-\dfrac{b^2}{2}+182$

$13-182=-\dfrac{b^2}{4}$

$-169=-\dfrac{b^2}{4}$

$169\times 4=b^2$

Taking square root on both sides.

$13\times 2=b$

$b=26$

The value of b is 26.

So, the given function is  

$f(x)=x^2+26x+182$

Now, add and subtract square of half of coefficient of x.

$f(x)=x^2+26x+182+(13)^2-(13)^2$

$f(x)=(x^2+2(13)x+(13)^2)+182-169$

$f(x)=(x+13)^2+13$

On comparing with $f(x)=(x-h)^2+k$, we get

$h=-13$

$k=13$

Therefore, your answer is incorrect.