The parabola with equation y=ax^2+bx+c is graphed below:
The zeros of the quadratic ax^2 + bx + c are at x=m and x=n, where m>n. What is m-n
?
+9479 Here's one way to solve this:
The equation of this parabola in vertex form is
y = a(x - 2)2 - 4 , where a is some constant that "stretches" or "shrinks" the parabola.
To find a we can use the fact that the parabola passes through the point (4, 12)
So we know the following must be true:
12 = a(4 - 2)2 - 4
Simplify the right side of the equation....
12 = 4a - 4
Add 4 to both sides of the equation.
16 = 4a
Divide both sides of the equation by 4
4 = a
So the equation of this parabola is
y = 4(x - 2)2 - 4
To find the zeros, we set y equal to 0 and solve for x
0 = 4(x - 2)2 - 4
Add 4 to both sides of the equation.
4 = 4(x - 2)2
Divide both sides of the equation by 4
1 = (x - 2)2
Take the ± square root of both sides.
±1 = x - 2
Split into two equations and solve for x ...
x - 2 = 1 or x - 2 = -1
x = 3 x = 1
So the zeros of the quadratic are at x = 3 and x = 1
3 - 1 = 2
+9479 Here's one way to solve this:
The equation of this parabola in vertex form is
y = a(x - 2)2 - 4 , where a is some constant that "stretches" or "shrinks" the parabola.
To find a we can use the fact that the parabola passes through the point (4, 12)
So we know the following must be true:
12 = a(4 - 2)2 - 4
Simplify the right side of the equation....
12 = 4a - 4
Add 4 to both sides of the equation.
16 = 4a
Divide both sides of the equation by 4
4 = a
So the equation of this parabola is
y = 4(x - 2)2 - 4
To find the zeros, we set y equal to 0 and solve for x
0 = 4(x - 2)2 - 4
Add 4 to both sides of the equation.
4 = 4(x - 2)2
Divide both sides of the equation by 4
1 = (x - 2)2
Take the ± square root of both sides.
±1 = x - 2
Split into two equations and solve for x ...
x - 2 = 1 or x - 2 = -1
x = 3 x = 1
So the zeros of the quadratic are at x = 3 and x = 1
3 - 1 = 2
