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

2 Answers
Apr 3, 2017

#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]}

Apr 4, 2017

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]}