Quadratic equation -

Question

I have been struggling with this quadratic equation question quite a bit. Specify the zero points for the following function and the coordinates of the parabol peak.

$$g(x)=4x^2 +36$$

Edit 1 - this is where I got stuck: $$g(x)=4(x^2+9)$$

Answer 1

HINT

To find the zero set

$$g(x)=4x^2 +36=0\implies x^2=-\frac{36}4=-9 $$

thus we don't have real solution, that means we don't have real roots for $g$.

For the peak let consider

$$g'(x)=8x=0\implies x=0$$

thus the peak is $g(0)=36$ which of course is a minimum.

Answer 2

Hint: $\;x^2+9\ge 9$ if $x$ is real.

Answer 3

If you don't know calculus, you can derive the maximum or minimum of a quadratic $$f(x)=ax^2+bx+c$$ by looking at it in vertex form: $$f(x)=a(x-h)^2+k,$$ where $h=\frac{-b}{2}$ and so $f(h)=k$ is the maximum or minimum value of this function, since it is the unique point which lies on the axis of symmetry.