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: $$g(x)=4(x^2+9)$$
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$\begingroup$ What did you try? $\endgroup$– Parcly TaxelFeb 10, 2018 at 0:33
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1$\begingroup$ @JamesLeaf: What if you wrote it as $$g(x) = 4 \left(x^2+9\right)$$ Hint: you get two complex conjugate imaginary roots. $\endgroup$– MooFeb 10, 2018 at 0:34
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$\begingroup$ Yeah i actually did that, but that is where i got stuck. $\endgroup$– James LeafFeb 10, 2018 at 0:34
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$\begingroup$ Didn't you set it equal to zero? $\endgroup$– saulspatzFeb 10, 2018 at 0:36
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$\begingroup$ What you call a peak looks very much like a glen… $\endgroup$– BernardFeb 10, 2018 at 0:37
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3 Answers
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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.
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Hint: $\;x^2+9\ge 9$ if $x$ is real.
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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.
