Question 120433
Given: the parabola
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{{{y = x^2-4}}}
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To find the intersections with the x-axis, set y = 0 because any point on the x-axis has
zero as its y-value. So the equation to solve then becomes:
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{{{0 = x^2-4}}}
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Solve this by getting rid of the -4 on the right side by adding +4 to both sides. On the right
side the -4 and the +4 cancel each other when they are added. So when you add +4 to both sides
the equation becomes:
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{{{4 = x^2}}}
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Solve for x by taking the square root of both sides to get two answers:
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{{{x = sqrt(4)=2}}} and {{{x = sqrt(4) = -2}}}
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So the intersections of the parabola and the x-axis occur on the x-axis at x = -2 and at
x = +2. These points are P and Q. And, if you subtract them you get:
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{{{2-(-2) = 2+2 = 4}}}
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The points P and Q are 4 units apart.
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You can see this in the graph of the parabola shown below:
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{{{graph(300,300,-5,5,-5,5,x^2 - 4)}}}
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Hope this helps you to understand the problem.