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: $$g(x)=4(x^2+9)$$
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: $$g(x)=4(x^2+9)$$
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.
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.