The general form of a parabola is given by the equation y = ax^2 + bx + c, where a is the coefficient of the x^2 term.
For the first parabola, we know that it crosses the x-axis at (-4, 0) and (6, 0), so the roots of the equation are -4 and 6. Therefore, the equation of the first parabola can be written as y = a(x + 4)(x - 6) = a(x^2 - 2x - 24).
Given that the y-intercept is at (0, -12), we can plug in the coordinates to solve for a:
-12 = a(0^2 - 2(0) - 24)
-12 = -24a
a = 1/2
For the second parabola, following the same process, we can write its equation as y = b(x + 4)(x - 6) = b(x^2 - 2x - 24).
Given that the y-intercept is at (0, -24), we can solve for b:
-24 = b(0^2 - 2(0) - 24)
-24 = -24b
b = 1
Therefore, the positive difference between the two values of a is:
|1/2 - 1| = 1/2
Rounded to the nearest tenth, the positive difference between the two a values is 0.5.
A student draws two parabolas on graph paper. Both parabolas cross the x-axis at (–4, 0) and (6, 0). The y-intercept of the first parabola is (0, –12). The y-intercept of the second parabola is (0, –24). What is the positive difference between the a values for the two functions that describe the parabolas? Write your answer as a decimal rounded to the nearest tenth.
1 answer