Question

y = ax^2+ bx + 9 has x-intercepts at x = -3 and x = 1.

what is the value of a?
what is the value of b?
please help!!

Answer 1

knowing the roots, you know that

a(x+3)(x-1) = ax^2+bx+9
a(x^2+2x-3) = ax^2+bx+9
2a = b
-3a = 9

Answer 2

alternate way:

f(x) = ax^2 + bx + 9
f(-3) = 9a - 3b + 9 = 0 , 3a - b = -3
f(1) = a + b = -9

add, 4a = -12
a = -3, mentally get b = -6

Answer 3

[AI-generated answer — Explain Bot]

To find the values of a and b in the equation y = ax^2 + bx + 9, we can use the given information about the x-intercepts.

1. Start with the x-intercepts:
The x-intercepts are the values of x where y = 0. In this case, the x-intercepts are x = -3 and x = 1.

2. Plug in the x-intercepts:
Since y = 0 at the x-intercepts, we can substitute the x-values into the equation:
When x = -3:
0 = a(-3)^2 + b(-3) + 9
0 = 9a - 3b + 9 (equation 1)

When x = 1:
0 = a(1)^2 + b(1) + 9
0 = a + b + 9 (equation 2)

3. Solve the system of equations:
Now we have a system of two equations, equation 1 and equation 2. We can solve this system to find the values of a and b.

Multiplying equation 2 by -3, we get:
0 = -3a - 3b - 27 (equation 3)

Now we can add equation 1 and equation 3:
0 = 9a - 3b + 9
0 = -3a - 3b - 27
-----------------------
0 = 6a - 36

4. Simplify and solve for a:
6a - 36 = 0
6a = 36
a = 36/6
a = 6

Therefore, the value of a is 6.

5. Substitute the value of a into equation 2 to find b:
0 = a + b + 9
0 = 6 + b + 9
0 = b + 15
b = -15

Therefore, the value of b is -15.

In conclusion, the value of a is 6 and the value of b is -15 in the equation y = ax^2 + bx + 9.