Question 675504
The standard quadratic equation is:
{{{y = ax^2+bx+c}}}
In this problem, x will represent the time and y will represent the height.<br>
The "c" in this equation is the y-intercept. Since the ball starts (x=0) on the ground (y = 0), c is 0. So our equation will have the form:
{{{y = ax^2+bx}}}
Now all we have to do is figure out the values for "a" and "b".<br>
The problem tells us that at 2 seconds the ball is 6 feet in the air. So the point (2, 6) is on this parabola and must fit our equation:
{{{6 = a*(2)^2 + b*2}}}
We are also told that after 4 seconds the ball is 8 feet in the air. So the point (4, 8) must also be on the parabola and fit our equation:
{{{8 = a*(4)^2 + b*4}}}
With these two equations we should be able to solve for a and b.<br>
However, there is a slightly faster way to find a and b. We are told that the height of 8 is the maximum height. This means that (4, 8) is not just any point on the parabola. It is the vertex of the (downward-opening) parabola. This fact can be exploited to find a and b a little faster than solving the two-equation system above. One way to take advantage of knowing the vertex uses the fact that the x-coordinate of the vertex of a parabola is always -b/2a. So
{{{4 = -b/2a)
If we multiply this by 2a we get:
8a = -b
or
-8a = b
We can use this equation and the {{{6 = a*(2)^2 + b*2}}} equation to find a and b. This system is easier and faster to solve since one equation, -8a = b, is already solved for b. Substituting for b in the other equation we get:
{{{6 = a*(2)^2 + (-8a)*2}}}
Simplifying...
{{{6 = a*4 + (-8a)*2}}}
{{{6 = 4a + (-16a)}}}
{{{6 = -12a}}}
Dividing by -12:
{{{-1/2 = a}}}
Now we can find b:
{{{b = -8a}}}
{{{b = -8(-1/2)}}}
{{{b = 4}}}
With a = -1/2 and b = 4 our equation is:
{{{y = (-1/2)x^2+4x}}}<br>
Now we can use this to find the height at 7 seconds:
{{{y = (-1/2)(7)^2+4(7)}}}
{{{y = (-1/2)49+4(7)}}}
{{{y = -49/2+28}}}
{{{y = -49/2+56/2}}}
{{{y = 9/2 = 4&1/2}}}
So the ball will be {{{4&1/2}}} feet off the ground after 7 seconds.<br>
Notes:<ul><li>You get the same values for a and b if you solve the original system of:
{{{6 = a*(2)^2 + b*2}}}
and
{{{8 = a*(4)^2 + b*4}}}</li><li>If you don't know (or can't remember) that the x-coordinate of the vertex is -b/2a there is another way to take advantage of the vertex. Parabolas are symmetric about the vertical line through the vertex. Since it took 4 seconds to get from the ground to the highest point it will take another 4 seconds to come back to the ground. So (8, 0) will be another point on this parabola. With the two x-intercepts of 0 (from (0, 0)) and 8 (from (8, 0)) we can write the equation of the parabola in a different way:
{{{y = k*(x-0)(x-8)}}}
Simplifying...
{{{y = k*x*(x-8)}}}
{{{y = k*x^2-8kx}}}
And we can use with (2, 6) or (4, 8) to find k:
{{{6 = k*(2)^2-8k(2))}}}
{{{6 = k*4-8k(2)}}}
{{{6 = k*4-16k}}}
{{{6 = -12k}}}
{{{-1/2 = k}}}
Substituting this back into:
{{{y = k*x^2-8kx}}}
{{{y = (-1/2)*x^2-8(-1/2)x}}}
{{{y = (-1/2)*x^2+4x}}}</li></ul>