Question

Use the zeros of the following quadratic to find the x-value of the vertex.

y = x^2 – 4x – 5
A.
5
B.
2
C.
–2
D.
–5

Answer 1

Lol , Im going to help you even though we are sitting next to each other !

Answer 2

Vertex = 2

Answer 3

when we are solving the quadratic equations then we have to take it equals to zero.

that is we have to take 0 = x^2 – 4x – 5

now we have to solve this for x.

we use here AC method of factoring.

The AC method of factoring is used to factor polynomials of the form
Ax² + Bx + C. A, B, and C represent constants and x is the variable.
Determine AC by multiplying the A term and C term.

x^2 – 4x – 5

by comparing we get A = 1 B = - 4 and C = -5

so the answer is A .

Find two numbers that add to the B term and multiply to AC.

that is find two numbers that add to -4 and multiply to AC that is -5*1=-5

so here we get -5+1 = -4 and
when we multiply them we get -5*1 = -5

Call the smaller number M and the larger number N.
so the smaller number is -5 and larger number = 1

Rewrite the original equation as Ax² + Mx + Nx + C.

that is x^2 -5x +x -5

now factor out the common term from first two terms.

here x is the common term in first two terms
so that we get x(x - 5) + x - 5

also find the common factor from last two terms

here 1 is common factor from last two terms.

so that we get x (x-5) +1(x - 5)

now we apply the distributive property.

(x+1)(x-5)

we get this factors. in the original problm
we have 0 = x^2-4x-5

we get right side is (x+1)(x-5)

so that 0 = (x+1)(x-5)

so we write it like (x+1) = 0 or (x-5) = 0

from first factor we get x = -1 and x = 5