# How do you solve the equation x^2-2x-24=0 by graphing?

Apr 3, 2017

$x = - 4 \text{ }$ or $\text{ } x = 6$

#### Explanation:

Given:

${x}^{2} - 2 x - 24 = 0$

I notice that the question asks to graph to solve. I would usually do the reverse: solve to graph, but let's have a look...

Let:

$f \left(x\right) = {x}^{2} - 2 x - 24$

Evaluating for a few values we find:

$f \left(0\right) = {\textcolor{b l u e}{0}}^{2} - 2 \left(\textcolor{b l u e}{0}\right) - 24 = 0 - 0 - 24 = - 24$

$f \left(1\right) = {\textcolor{b l u e}{1}}^{2} - 2 \left(\textcolor{b l u e}{1}\right) - 24 = 1 - 2 - 24 = - 25$

$f \left(2\right) = {\textcolor{b l u e}{2}}^{2} - 2 \left(\textcolor{b l u e}{2}\right) - 24 = 4 - 4 - 24 = - 24$

Interesting!

Notice that $f \left(0\right) = f \left(2\right)$

Since this is a quadratic in $x$ and these two values are equal, then the points $\left(0 , - 24\right)$ and $\left(2 , - 24\right)$ are at equal distance from the axis, which must be the line $x = 1$, running through the vertex $\left(1 , - 25\right)$.

At this point we could note that the multiplier of the ${x}^{2}$ term is $1$, so this quadratic is just like $y = {x}^{2}$, but with the vertex translated to $\left(1 , - 25\right) = \left(1 , - {5}^{2}\right)$. Then we could deduce that the $x$ intercepts must be at $x = 1 \pm 5$, giving us our two solutions.

We can check our deduction:

$f \left(- 4\right) = {\left(\textcolor{b l u e}{- 4}\right)}^{2} - 2 \left(\textcolor{b l u e}{- 4}\right) - 24 = 16 + 8 - 24 = 0$

$f \left(6\right) = {\textcolor{b l u e}{6}}^{2} - 2 \left(\textcolor{b l u e}{6}\right) - 24 = 36 - 12 - 24 = 0$

Here's the actual graph, with some of the features we have deduced:

graph{(y-(x^2-2x-24))(x-1+0.0001y)(10(x-1)^2+(y+25)^2-0.1)(10x^2+(y+24)^2-0.1)(10(x-2)^2+(y+24)^2-0.1)(10(x+4)^2+y^2-0.1)(10(x-6)^2+y^2-0.1) = 0 [-10, 10, -30, 15]}

Apr 4, 2017

An algebraic solution is not asked for.

Draw the graph and read the values of the $x$-intercepts.

$x = - 4 \mathmr{and} x = 6$

#### Explanation:

The first step is to draw the graph of $y = {x}^{2} - 2 x - 24$

You can do this by choosing several $x$ values and then finding the $y$ value for each. About $7$ points is a good number to use.
Plot the points and draw the parabola.

You could also find the significant points by calculation:
The $y -$intercept
The axis of symmetry and hence the turning point.
The $x$-intercepts

Once you have the graph, you can turn your attention to answering the question:

solve ${x}^{2} - 2 x - 24 = 0$ from a graph.

If you compare the equations:
$y = {x}^{2} - 2 x - 24 \text{ and } 0 = {x}^{2} - 2 x - 24$,

you will realise that $y = 0$

The question being asked is "Where does the parabola cross the $x$-axis?"

You can read these values as the $x$-intercepts from the graph.

These are seen to be $x = - 4 \mathmr{and} x = 6$
graph{y= x^2 -2x -24 [-10, 10, -5, 5]}