Find the quadratic function that has a vertex at (3,4) and goes through the point (0,13).
f(x)=
the vertex tells you that
y = a(x-3)^2 + 4
Now plug in (0,13) to find a
13 = a(0-3)^2 + 4
To find the quadratic function, we can use the vertex form of a quadratic equation, which is:
f(x) = a(x - h)^2 + k
where (h, k) is the vertex of the parabola.
Given that the vertex is (3, 4), we have h = 3 and k = 4.
So, the equation becomes:
f(x) = a(x - 3)^2 + 4
To find the value of 'a', we need to substitute the coordinates of the other given point (0, 13) into the equation:
13 = a(0 - 3)^2 + 4
13 = 9a + 4
9a = 13 - 4
9a = 9
a = 1
So the quadratic function that has a vertex at (3, 4) and goes through the point (0, 13) is:
f(x) = (x - 3)^2 + 4
To find the quadratic function that has a vertex at (3,4) and goes through the point (0,13), we can start by using the vertex form of a quadratic function:
f(x) = a(x - h)^2 + k
Where (h,k) represents the coordinates of the vertex.
Given that the vertex is (3,4), we can substitute these values into the equation:
f(x) = a(x - 3)^2 + 4
Next, we need to find the value of 'a' to complete the equation. To do this, we can use the point (0,13) that the quadratic function passes through. By substituting these values into the equation, we can solve for 'a':
13 = a(0 - 3)^2 + 4
Simplifying this equation, we get:
13 = 9a + 4
Subtracting 4 from both sides of the equation, we have:
9 = 9a
Dividing both sides by 9, we find:
a = 1
Now we can write the final quadratic function by substituting 'a' into the equation:
f(x) = 1(x - 3)^2 + 4
Therefore, the quadratic function that has a vertex at (3,4) and goes through the point (0,13) is:
f(x) = (x - 3)^2 + 4