A quadratic function has its vertex at the point (-9,3). The function passes through the point (2,-1). Find the quadratic and linear coefficients and the constant term of the function.
To find the quadratic function, we need to use the vertex form of a quadratic function, which is given by:
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the quadratic function.
Given that the vertex is (-9, 3), we can substitute these values into the equation:
f(x) = a(x - (-9))^2 + 3
Simplifying this equation, we have:
f(x) = a(x + 9)^2 + 3
Now, we need to find the value of 'a'. To do this, we can use the other given point (2, -1). By substituting the coordinates into the equation, we can solve for 'a':
-1 = a(2 + 9)^2 + 3
-1 = a(11)^2 + 3
-4 = a(121)
a = -4/121
Now that we have the value of 'a', we can substitute it back into the equation to find the quadratic function:
f(x) = (-4/121)(x + 9)^2 + 3
The quadratic coefficient is -4/121. The linear coefficient is 0 since there is no term with just 'x'. The constant term is 3.
To find the quadratic and linear coefficients, and the constant term of the quadratic function, you need to use the information about the vertex and a point it passes through.
Let's denote the quadratic function as f(x) = ax^2 + bx + c, where a, b, and c are the coefficients we need to find.
Given that the vertex of the quadratic function is (-9,3), we can note that the x-coordinate of the vertex (-9) gives us the value of x when the function has its maximum or minimum value. In this case, the vertex tells us that the function has a minimum value because the coefficient of x^2 is positive.
Using the coordinates of the vertex, we can determine that x = -9 and y = 3 satisfy the quadratic equation:
3 = a(-9)^2 + b(-9) + c
3 = 81a - 9b + c (equation 1)
Additionally, we are given that the quadratic function passes through the point (2,-1). Plugging these coordinates into the quadratic equation, we get:
-1 = a(2)^2 + b(2) + c
-1 = 4a + 2b + c (equation 2)
You now have a system of two equations with three unknowns (a, b, and c).
To solve this system, we can use substitution or elimination method. I will use the elimination method.
Subtracting equation 2 from equation 1, we get:
3 - (-1) = (81a - 4a) + (-9b - 2b) + (c - c)
4 = 77a - 11b (equation 3)
Now, we have two equations:
77a - 11b = 4 (equation 3)
3 = 81a - 9b + c (equation 1)
We can solve equation 3 for a in terms of b:
77a = 4 + 11b
a = (4 + 11b) / 77
Substituting this value of a into equation 1:
3 = 81[(4 + 11b) / 77] - 9b + c
To simplify this, we can multiply through by 77 to get rid of the fraction:
231 = 4(4 + 11b) - 9b(77) + 77c
231 = 16 + 44b - 693b + 77c
231 = -677b + 16 + 77c
215 = -677b + 77c
Now we have an equation relating b and c. We can choose values for b or c and solve for the other. Let's assume b = 1:
215 = -677(1) + 77c
215 = -677 + 77c
892 = 77c
c = 892 / 77
Now that we have the value of c, we can substitute it back into equation 3 to get a:
77a - 11(1) = 4
77a - 11 = 4
77a = 4 + 11
77a = 15
a = 15 / 77
So, the quadratic function in vertex form is:
f(x) = (15 / 77)x^2 + x + (892 / 77)
Therefore, the quadratic coefficient is 15/77, the linear coefficient is 1, and the constant term is 892/77.
let the function be
y = a(x+9)^2 + 3
sub in (2,-1) to find a , and you are done.