Find expressions for the quadratic functions whose graphs are shown. One graph has the point (4,2) plotted in which the parabola passes through (U-shaped parabola- right side up) The vertex is at (3,0) and the parabola does not touch the y-axis for as much is shown.
So, I did this:
a(3)^2 + b(3) + c = 0
Solve: c = -12
For the point (4,2):
a(4)^2 + b(4) - 12 = 2
Solve: 16a + 4b = 14
I am not sure if I am doing this correctly and I am also stuck at the is point. Please help.
Your approach is partially correct so far!
To find the quadratic function that represents the given graph, we can use the general form of a quadratic function: f(x) = ax^2 + bx + c, where a, b, and c are constants.
From the given information, we can determine that the vertex form of the quadratic function is f(x) = a(x - h)^2 + k, where (h, k) represents the vertex. In this case, the vertex is (3, 0), so the quadratic function can be written as f(x) = a(x - 3)^2 + 0, which simplifies to f(x) = a(x - 3)^2.
To find the constant 'a', we can use the point (4, 2) that lies on the graph of the quadratic function. Plugging these coordinates into the equation, we have:
2 = a(4 - 3)^2
2 = a(1)^2
2 = a * 1
a = 2
Thus, the quadratic function that represents the given graph is f(x) = 2(x - 3)^2.
Now, to specifically address the equation you set up:
- You correctly set up the equation based on the vertex form: a(3)^2 + b(3) + c = 0.
- However, the equation is used to determine the value of 'c', which represents the y-coordinate of the vertex. In this case, you correctly solved for c = -12.
- While it is a good practice to check the equation using a point on the graph, the given point (4, 2) is not sufficient to determine the values of 'a' and 'b' using this equation.
To find 'a' and 'b', you need to use the point-slope form of the quadratic function and solve a system of equations. Since the y-coordinate of the vertex is 0 and the parabola does not touch the y-axis as much is shown, we have:
c = -12 (from the equation a(3)^2 + b(3) + c = 0)
16a + 4b = 14 (from the point (4, 2))
To solve this system, substitute c = -12 into the second equation:
16a + 4b = 14
16a + 4b = 14 - 12
16a + 4b = 2
Now, you can solve for 'a' and 'b' using any method such as elimination, substitution, or matrices.
You are on the right track! Let's continue solving the system of equations to find the quadratic function.
From your first equation, we have:
a(3)^2 + b(3) - 12 = 0
Simplifying this equation gives:
9a + 3b - 12 = 0
Now, let's use the second equation using the point (4,2):
a(4)^2 + b(4) - 12 = 2
Simplifying this equation gives:
16a + 4b - 12 = 2
16a + 4b = 14
Now, we have a system of two equations:
9a + 3b - 12 = 0
16a + 4b = 14
We can solve this system of equations using the method of substitution or elimination. I'll use the method of substitution here.
First, let's solve the first equation for a in terms of b:
9a = 12 - 3b
a = (12 - 3b) / 9
a = (4 - b) / 3
Substituting this expression for a into the second equation, we have:
16((4 - b) / 3) + 4b = 14
Multiplying everything by 3 to eliminate the fractions, we get:
16(4 - b) + 12b = 42
64 - 16b + 12b = 42
-4b + 64 = 42
-4b = 42 - 64
-4b = -22
b = -22 / -4
b = 11/2 or 5.5
Now, substitute the value of b back into the expression for a:
a = (4 - (11/2)) / 3
a = (8 - 11) / 6
a = -3/6
a = -1/2 or -0.5
So, the quadratic function can be written as:
f(x) = -0.5x^2 + 5.5x - 12
Therefore, the expression for the quadratic function whose graph passes through the point (4,2) and has the given vertex is f(x) = -0.5x^2 + 5.5x - 12.
c does not equal -12:
a(3)^2 + b(3) + c = 0
Solve: c = -9a - 3b (not -12)
For the given parabola, since the vertex is (3,0), and knowing that parabolas are symmetrical with respect to the vertex, we therefore have another point (2,2), the mirror image of (4,2) about x=3 passing through the vertex.
The three known points give rise to three equations with three unknowns (a,b,c). Solve for a,b and c.