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Th equation of the parabola is

$$2\left(x+\dfrac34\right)^2−\dfrac{25}8$$

What is the vertex and the min value? and do I just plug $x$ values into the equation to get the points on the graph?

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The vertex will be where the squared part equals zero as that is where the axis of symmetry is for the parabola. As for the minimum value, consider a few values around the vertex and you should notice a pattern, I think. –  JB King Oct 13 '13 at 18:31
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Let be $f(x)=2\left(x+\frac34\right)^2−\frac{25}8$. If you expand squared term

$$f(x)=2\left(x+\frac34\right)^2−\frac{25}8=2x^2+3x+\frac{9}{8}-\frac{25}{8}$$ $$f(x)=2x^2+3x+\frac{9}{8}-\frac{25}{8}=2x^2+3x-2$$

the abscissa of vertex of $f$ is $x$ such that $f'(x)=0$. So $f'(x)=4x+3=0 \Rightarrow x=\frac{-3}{4}$. If you substitute $x=\frac{-3}{4}$ into $f(x)$ you will find ordinate of vertex as $\frac{-25}{8}$ which also is minimum of $f$. So the minimum is $\frac{-25}{8}$ and the coordinates of vertex are $$\left( \frac{-3}{4},\frac{-25}{8}\right)$$


If you are not familiar with derivative you can do the things as follows: Let us denote coordinates of vertex by $T(r,k)$. Recall if $f(x)=ax^2+bx+c$ then $r=\frac{-b}{2a}$ and $k=f(r)$. Also recall $k=f(r)$ is extremum of $f$. Now in our case $f(x)=2x^2+3x-2$ and $a=2,b=3,c=-2$.Hence $$r=\frac{-b}{2a}=\frac{-3}{2.2}=\frac{-3}{4}$$ $$k=f(r)=f(\frac{-3}{4})=\frac{-25}{8}$$ $$T(r,k)=\left( \frac{-3}{4},\frac{-25}{8}\right)$$

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