Question

Find the maximum or minimum value of the quadratic expression $2x-7-5x^2$.

Answer 1

Let $y=2x-7-5x^2$

$\Rightarrow y^{\prime }=2-10x=0$, for max/min

$\Rightarrow x=\frac{1}{5}$

Also $y^{\prime \prime }=-10<0$

so maximum value of $y$ is $=y\left(\frac{1}{5}\right)=2\cdot \frac{1}{5}-7-5\cdot \frac{1}{25}=\frac{2}{5}-7-\frac{1}{5}=\frac{1}{5}-7=-\frac{34}{7}$

And minimum value of $y$ will be $-\infty$