What is the vertex of y= 2x^2 +5x + 30?

Jun 18, 2017

The vertex of $y$ is the point $\left(- 1.25 , 26.875\right)$

Explanation:

For a parabola in standard form: $y = a {x}^{2} + b x + c$

the vertex is the point where $x = \frac{- b}{2 a}$

NB: This point will be a maximum or minimum of $y$ depending on the sign of $a$

In our example: $y = 2 {x}^{2} + 5 x + 30 \to a = 2 , b = 5 , c = 30$

$\therefore {x}_{\text{vertex}} = \frac{- 5}{2 \times 2}$

$= - \frac{5}{4} = - 1.25$

Replacing for $x$ in $y$

${y}_{\text{vertex}} = 2 \times {\left(- \frac{5}{4}\right)}^{2} + 5 \times \left(- \frac{5}{4}\right) + 30$

$= 2 \times \frac{25}{16} - \frac{25}{4} + 30$

$= \frac{50}{16} - \frac{100}{16} + 30 = - \frac{50}{16} + 30$

$= 26.875$

The vertex of $y$ is the point $\left(- 1.25 , 26.875\right)$

We can see this point as the minimum of $y$ on the graph below.
graph{2x^2+5x+30 [-43.26, 73.74, -9.2, 49.34]}

Jun 18, 2017

To find the vertex, the easiest thing to do (besides graphing the problem) is to convert the equation into vertex form. To do that, we should "complete the square"

$y = 2 {x}^{2} + 5 x + 30$

the leading coefficient must be $1$, so factor out the $2$

$y = 2 \left({x}^{2} + \frac{5}{2} x + 6\right)$

We need to find a value that changes ${x}^{2} + \frac{5}{2} x + 6$ into a perfect square.

To do that, we need to take the middle term, $\frac{5}{2}$, and divide it by $2$. That gives us $\frac{5}{4}$.

Our next step is to square the result: ${\left(\frac{5}{4}\right)}^{2}$, or $\frac{25}{16}$

$- - - - - - - - - - - - - -$

Now we have our missing value: ${x}^{2} + \frac{5}{2} x + 6 + \frac{25}{16}$ WAIT We can't just add something to a problem! But, if we add something and then immediately subtract it, technically we haven't changed the equation, since they subtract out to zero

So, our problem really is ${x}^{2} + \frac{5}{2} x + 6 + \frac{25}{16} - \frac{25}{16}$

Let's rewrite this: ${x}^{2} + \frac{5}{2} x + \frac{25}{16} + 6 - \frac{25}{16}$

${x}^{2} + \frac{5}{2} x + \frac{25}{16}$ is a perfect square. Let's rewrite it in that form: ${\left(x + \frac{5}{4}\right)}^{2}$

Now let's look at our equation again: ${\left(x + \frac{5}{4}\right)}^{2} + 6 - \frac{25}{16}$

Let's combine like-terms: ${\left(x + \frac{5}{4}\right)}^{2} + \frac{71}{16}$

Now we have the equation in vertex form, and we can find the vertex very easily from here

${\left(x + \textcolor{red}{\frac{5}{4}}\right)}^{2} + \textcolor{y e l l o w}{\frac{71}{16}}$

$\left(- \textcolor{red}{x} , \textcolor{y e l l o w}{y}\right)$

$\left(- \textcolor{red}{\frac{5}{4}} , \textcolor{y e l l o w}{\frac{71}{16}}\right)$

That's the vertex.

To check our work, let's graph our equation and see the vertex
graph{y=2x^2+5x+30}

We were right! $- 1.25$ and $4.4375$ are equivalent to $- \frac{25}{16}$ and $\frac{71}{16}$