# How do you graph the parabola f(x)=x ^2 + 3x-10 using vertex, intercepts and additional points?

Jan 11, 2018

After calculating the vertex and intercepts and plotting them, calculate additional points, $\left(x , f \left(x\right)\right)$, near these crucial spots of our parabola.

#### Explanation:

To find the vertex, we use the formula $x = - \frac{b}{2 a}$, but as you may have guessed, this will only give us the $x$ coordinate of the vertex. In order to get the $y$ coordinate, we would need to put our $x$ value back into $f \left(x\right)$.

Standard form is
$a {x}^{2} + b x - 10$

In our case:
$a = 1$
$b = 3$

This means the $x$ coordinate of our vertex will be

$- \frac{3}{2 \left(1\right)}$ or simply $- \frac{3}{2}$

Plugging the $x$ coordinate back into $f \left(x\right)$, we get the $y$ coordinate.

${\left(- \frac{3}{2}\right)}^{2} + 3 \left(- \frac{3}{2}\right) - 10$

$= \frac{9}{4} - \frac{9}{2} - 10$

$= - \frac{49}{4}$

Our vertex is therefore $\left(- \frac{3}{2} , - \frac{49}{4}\right)$.

Now, we can calculate the $x$-intercept and the $y$-intercept, which, alongside our vertex, will both help us determine what other nearby points we may wish to calculate.

The $x$-intercept is where $y = 0$. This case may be solved by factoring.

${x}^{2} + 3 x - 10 = 0$

$\left(x + 5\right) \left(x - 2\right) = 0$

$x + 5 = 0 \mathmr{and} x - 2 = 0$

$x = - 5 \mathmr{and} x = 2$

So, our $x$-intercepts are $\left(- 5 , 0\right)$ and $\left(2 , 0\right)$.

The $y$-intercept is where $x = 0$. Simply plug this value into $f \left(x\right)$.

${\left(0\right)}^{2} + 3 \left(0\right) - 10$

$= - 10$

So, our $y$-intercept is $\left(0 , - 10\right)$.

Now, we have four relevant points that we can plot. These are:

• The vertex, $\left(- \frac{3}{2} , - \frac{49}{4}\right)$
• The $x$-intercepts, $\left(- 5 , 0\right)$ and $\left(2 , 0\right)$
• And the $y$-intercept, $\left(0 , - 10\right)$

After plotting these, we likely wish to plot some additional points in between to more accurately graph it. We would choose $x$ values and plug them into $f \left(x\right)$.

These would be points within our range of crucial coordinates, so anywhere between our lowest $x$ coordinate, $\left(- 5 , 0\right)$, and our greatest $x$ coordinate, $\left(2 , 0\right)$.

You may also plot more points outside of this range to give a more complete picture.

Jan 11, 2018

#### Explanation:

I have handwritten my explanation, I hope it is legible :)

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