How do you write a quadratic equation with x-intercepts: -1,-1 ; point: (2,-9) ?

Question

Eli Assouline · Accepted Answer

#y=-x^2-2x-1# #"given the x-intercepts (roots ) are " x=-1,x=-1# #"then the factors are " (x+1),(x+1)=(x+1)^2# #rArry=a(x+1)^2# #"to find a, substitute " (2,-9)" into the equation"# #rArr-9=a(2+1)^2# #rArr9a=-9rArra=-1# #rArry=-(x+1)^2larrcolor(red)" in intercept form"# #"expanding and simplifying gives"# #y=-x^2-2x-1larrcolor(red)" in standard form"# graph{-(x+1)^2 [-10, 10, -5, 5]}

Genesis Allen · Answer

undefined To write a quadratic equation with given x-intercepts and a point, you can use the factored form of a quadratic equation. Given x-intercepts at -1 and -1, the factors would be (x + 1) and (x + 1).

So, the factored form of the quadratic equation is (x + 1)(x + 1).

Next, use the given point (2, -9) to find the corresponding y-value when x = 2.

Substitute x = 2 into the equation and solve for the constant term:

(2 + 1)(2 + 1) = (3)(3) = 9

So, when x = 2, y = 9.

The given point has a y-value of -9. To make it match the given point, there needs to be a reflection over the x-axis, so the constant term in the equation should be -9.

Therefore, the quadratic equation is:

y = (x + 1)(x + 1) = (x + 1)^2 = x^2 + 2x + 1

Multiplying (x + 1)(x + 1) gives x^2 + 2x + 1.

So, the quadratic equation is y = x^2 + 2x - 9.