Question

Parabola vertex property Prove that if a parabola crosses the \(x\) -axis twice, the \(x\) -coordinate of the vertex of the parabola is halfway between the \(x\) -intercepts.

Short answer

Question: Prove that the x-coordinate of the vertex of a parabola is the midpoint of the x-intercepts. Answer: To prove this, we wrote the equation of the parabola in vertex form and found the x-intercepts. We then solved for the values of a and k in terms of x-intercepts and h, showing that the x-coordinate of the vertex (h) is the average of the x-intercepts: h = (x1+x2) / 2. This confirms that the x-coordinate of the vertex of the parabola is halfway between the x-intercepts.

Step-by-step solution

Write the parabola in vertex form

The equation of a parabola in vertex form is given by: $$ y = a(x-h)^2 + k $$ where \((h, k)\) is the vertex of the parabola, and \(a\) is a constant.

Find the x-intercepts of the parabola

To find the x-intercepts, set the y-value to 0 and solve for x: $$ 0 = a(x-h)^2 + k $$ Let's call the x-intercepts \(x_1\) and \(x_2\): $$ 0 = a(x_1-h)^2 + k \\ 0 = a(x_2-h)^2 + k $$

Solve for a and k in terms of x-intercepts and h

We'll now solve the system of equations for a and k in terms of \(x_1, x_2,\) and \(h\): $$ a(x_1-h)^2 = -k \\ a(x_2-h)^2 = -k $$ Dividing the two equations, we get: $$ \frac{(x_1-h)^2}{(x_2-h)^2} = 1 $$ Taking the square root of both sides: $$ \frac{x_1-h}{x_2-h} = \pm 1 $$

Solve for h in terms of x-intercepts

Now solve for \(h\): $$ x_1-h = \pm (x_2-h) $$ Add \(h\) to both sides and divide by 2 to get: $$ h = \frac{x_1 + x_2}{2} $$

Conclusion

We found that the x-coordinate of the vertex \(h\) is the average of the x-intercepts \(x_1\) and \(x_2\): $$ h = \frac{x_1+x_2}{2} $$ This proves that the x-coordinate of the vertex of the parabola is halfway between the x-intercepts.