How do you solve #x^2 - 16 = 0# graphically?

Question

Eli Assouline · Accepted Answer

On the graph below the x-intercepts are found at #x=-4 and x=+4#

You need to draw the graph of #y-x^2-16# first.

This can be done by plotting points which you work out using the equation. Choose several x-values... negative, zero and positive, then calculate the corresponding y-values.

The graph is a parabola - draw a smooth, free-hand curve through all the points.

Now you can solve the equation graphically.

Compare the equation of the GRAPH with the EQUATION to be solved.

#x^2 -16 = color(red)(y) and x^2-16 = color(red)(0)# This gives #color(red)(y=0)#

#color(red)(y=0)# is the equation of the #color(red)(x)#- axis.

The question is actually asking..

"Where does the parabola cross the x-axis?"

OR:

"What are the points of intersection of the two graphs #y=x^2-16 and y=0#?"

On the graph below the x-intercepts are found at #x=-4 and x=+4#

(Note that the y-intercept is at #y=-16#) graph{y=x^2-16 [-9.79, 10.21, -5.88, 4.12]}

Ethan Adkins · Answer

undefined To solve $x^2 - 16 = 0$ graphically, you would plot the quadratic function $y = x^2 - 16$ on a graph. Then, you would identify the points where the graph intersects the x-axis, as these are the solutions to the equation. In this case, the graph intersects the x-axis at $x = -4$ and $x = 4$, so the solutions to the equation $x^2 - 16 = 0$ are $x = -4$ and $x = 4$.