Question

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

Answer 1

`x=-4" "` or `" "x=6`

Explanation:

Given:

`x^2-2x-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(x) = x^2-2x-24`

Evaluating for a few values we find:

`f(0) = color(blue)(0)^2-2(color(blue)(0))-24 = 0-0-24 = -24`

`f(1) = color(blue)(1)^2-2(color(blue)(1))-24 = 1-2-24 = -25`

`f(2) = color(blue)(2)^2-2(color(blue)(2))-24 = 4-4-24 = -24`

Interesting!

Notice that `f(0) = f(2)`

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

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 `(1, -25) = (1, -5^2)`. Then we could deduce that the `x` intercepts must be at `x = 1+-5`, giving us our two solutions.

We can check our deduction:

`f(-4) = (color(blue)(-4))^2-2(color(blue)(-4))-24 = 16+8-24 = 0`

`f(6) = color(blue)(6)^2-2(color(blue)(6))-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]}

Answer 2

An algebraic solution is not asked for.

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

`x = -4 and x=6`

Explanation:

The first step is to draw the graph of `y = x^2 -2x -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 -2x -24 =0` from a graph.

If you compare the equations:
`y = x^2 -2x -24" and " 0 = x^2 -2x -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 and x=6`
graph{y= x^2 -2x -24 [-10, 10, -5, 5]}