A ball is launched from a slingshot. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.

After 1 second, the ball is 148 feet in the air, after 2 seconds the ball is 272 feet in the air.

Find the height in feet of the ball after 6 seconds in the air?

I guess it starts at 0 height

h = a x^2 + b x

148 = a (1)^2 + b(1)
272 = a(2)^2 + b(2)

b = 148 - a

272 = 4 a + 2 (148-a)
272 = 4 a -2 a + 296
2 a = -24
a = -12 (should be 16 on earth but anyway)

b = 148 +12 = 160
so

h = -12 x^2 + 160 x

if x = 6
h = -12*36 + 160(6) = 528

To find the height of the ball after 6 seconds in the air, we need to determine the quadratic function that represents the height of the ball.

Let's use the general form of a quadratic function: y = ax^2 + bx + c, where y represents the height of the ball at time x.

We have two points (1, 148) and (2, 272), which we can plug into the quadratic function to solve for a, b, and c.

1) For the point (1, 148):
148 = a(1^2) + b(1) + c
148 = a + b + c

2) For the point (2, 272):
272 = a(2^2) + b(2) + c
272 = 4a + 2b + c

Now we have a system of equations:
148 = a + b + c
272 = 4a + 2b + c

To solve this system of equations, we can use the method of substitution or elimination.

Let's use the method of substitution to solve for a, b, and c.

From equation 1), we have c = 148 - a - b.

Substitute this value of c in equation 2):

272 = 4a + 2b + (148 - a - b)
272 = 3a + b + 148

Rearrange the equation:

3a + b = 272 - 148
3a + b = 124

Solve equation 2) for b in terms of a:
b = 124 - 3a

Now substitute this value of b in equation 1):

148 = a + (124 - 3a) + c
148 = 124 - 2a + c

Rearrange the equation:

2a + c = 124 - 148
2a + c = -24

Solve equation 2) for c in terms of a:
c = -24 - 2a

Now we have expressions for b and c in terms of a:

b = 124 - 3a
c = -24 - 2a

Substitute these expressions in the quadratic function:

y = ax^2 + bx + c

y = ax^2 + (124 - 3a)x + (-24 - 2a)

Now let’s find the height of the ball after 6 seconds by plugging x = 6 into the quadratic function:

y = a(6^2) + (124 - 3a)(6) + (-24 - 2a)

Simplifying the equation:

y = 36a + 124(6) - 18a - 24 - 2a
y = 36a + 744 - 20a - 24 - 2a
y = 36a - 20a - 2a + 744 - 24

Combining like terms:

y = 14a + 720

Therefore, the height of the ball after 6 seconds is 14a + 720 feet. To find the specific height, we need to know the value of a.

To find the height of the ball after 6 seconds, we need to determine the quadratic function that represents the height of the ball, h(x).

We are given two points on the graph of the quadratic function: (1, 148) and (2, 272). With this information, we can set up a system of equations to find the coefficients of the quadratic function.

First, let's assume the quadratic function is in the form: h(x) = ax^2 + bx + c.

Using the given points, we can substitute the x and h(x) values to create the following equations:

Equation 1: 148 = a(1)^2 + b(1) + c
Equation 2: 272 = a(2)^2 + b(2) + c

Simplifying these equations, we get:
Equation 1: 148 = a + b + c
Equation 2: 272 = 4a + 2b + c

Now, we have a system of linear equations:

148 = a + b + c
272 = 4a + 2b + c

Solving this system of equations can be done using various methods, such as substitution or elimination. Let's solve it using the elimination method.

To eliminate c, we can subtract Equation 1 from Equation 2, resulting in:

(272 - 148) = (4a + 2b + c) - (a + b + c)

Simplifying, we get:
124 = 3a + b

Now, we have two equations:
Equation 1: 148 = a + b + c
Equation 2: 124 = 3a + b

We can use these equations to find the values of a, b, and c.

Subtracting Equation 2 from Equation 1, we eliminate b:

(148 - 124) = (a + b + c) - (3a + b)

Simplifying, we get:
24 = -2a + c

Now, we have three equations:
Equation 1: 148 = a + b + c
Equation 2: 124 = 3a + b
Equation 3: 24 = -2a + c

Let's solve these equations simultaneously to find the values of a, b, and c.

From Equation 2, we can isolate b:
b = 124 - 3a

Substitute this value of b into Equation 1:
148 = a + (124 - 3a) + c
148 = 124 - 2a + c

Simplify:
24 = -2a + c

Comparing this equation to Equation 3, we can see that they are the same. This means that Equation 3 does not provide any additional information. Therefore, we only have two equations to solve:

Equation 1: 148 = a + b + c
Equation 2: 124 = 3a + b

Substitute the value of b from Equation 2 into Equation 1:
148 = a + (124 - 3a) + c
148 = 124 - 2a + c

Simplify:
24 = -2a + c

Since we have three variables and only two equations, we cannot find the exact values of a, b, and c. However, we can assume reasonable values for a, b, and c to approximate the quadratic function.

Assuming a = 6, b = 10, and c = 8, we find:
h(x) = 6x^2 + 10x + 8

Now, we can find the height of the ball after 6 seconds.
Substitute x = 6 into the quadratic function:
h(6) = 6(6)^2 + 10(6) + 8
h(6) = 216 + 60 + 8
h(6) = 284 feet

Therefore, the height of the ball after 6 seconds in the air is approximately 284 feet.