Answer:
y = 4x^2 + 32x + 28
Explanation:
Before we can find the standard form of the quadratic function with the given coordinates, we must first start with the intercept form, whose general equation is
y = a(x - p)(x - q), where
- a is a constant determining concavity (essentially, whether the parabola opens upward or downward)
- (x, y) are any point on the parabola,
- and p and q are the x-intercepts/roots
Step 1: We can plug in (-6, -20) for x and y, -7 for p and -1 for q into the intercept form. This will allows us to solve for a:
-20 = a(-6 - (-7))(-6 - (-1))
-20 = a(-6 + 7)(-6 + 1)
-20 = a(1)(-5)
-20 = -5a
4 = a
Thus, the full equation in vertex form is
y = 4(x + 7)(x + 1).
Step 1: The general equation for standard form is
y = ax^2 + bx + c.
We can convert from vertex to standard form by simply expanding the expression. Let's ignore the 4 for a moment simply focus on (x + 7)(x + 1).
We can expand using the FOIL method, where you multiply
- the first terms,
- the outer terms,
- the inner terms,
- and the last terms,
- then simplify by combining like terms
We see that the first terms are x and x, the outer terms are x and 1, the inner terms are 7 and x and the last terms are 7 * 1. Now, we multiply and simplify:
(x * x) + (x * 1) + (7 * x) + (7 * 1)
x^2 + x + 7x + 7
x^2 + 8x + 7
Step 3: Now, we can distribute the four to each term with multiplication:
4(x^2 + 8x + 7)
4x^2 + 32x + 28
Optional Step 4: We can check that our quadratic function in standard form, by plugging in -7, -1, and -6 for x and seeing that we get 0 as the y value for both x = -7 and x = -1 and -20 as the y value for x = -6:
Checking that (-7, 0) lies on the parabola of 4x^2 + 32x + 28:
0 = 4(-7)^2 + 32(-7) + 28
0 = 4(49) - 224 + 28
0 = 196 - 196
0 = 0
Checking that (-1, 0) lies on the parabola of 4x^2 + 32x + 28:
0 = 4(-1)^2 + 32(-1) + 28
0 = 4(1) - 32 + 28
0 = 4 - 4
0 = 0
Checking that (-6, -20) lies on the parabola of 4x^2 + 32x + 28:
-20 = 4(-6)^2 + 32(-6) + 28
-20 = 4(36) -192 + 28
-20 = 144 -164
-20 = -20
I attached a graph from Desmos to show how the function y = 4x^2 + 32x + 28 contains the points (-7, 0), (-1, 0), (-6, 20), further proving that we've correctly found the quadratic function in standard form passing through these three points