Answer:
(a) [tex](x-2)(x+4)[/tex]
zeros: x = 2, x = -4
vertex: (-1, -9)
(b) [tex]-(x+2)(x+7)[/tex]
zeros: x = -2, x = -7
vertex = (-4.5, 6.25)
Step-by-step explanation:
To factor a quadratic in the form [tex]ax^2+bx+c[/tex]:
- Find 2 two numbers (d and e) that multiply to [tex]ac[/tex] and sum to [tex]b[/tex]
- Rewrite [tex]b[/tex] as the sum of these 2 numbers: [tex]d+e=b[/tex]
- Factorize the first two terms and the last two terms separately, then factor out the comment term.
To find zeros of a factored quadratic in the form [tex](ax+b)(cx+d)[/tex]
- Set each of the parentheses to zero and solve for [tex]x[/tex]
The midpoint between the two zeros is the x-coordinate of the vertex. To find the y-coordinate of the vertex, substitute this into the given equation.
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Question (a)
Factored
Factor [tex]y=x^2+2x-8[/tex]
[tex]ac=-8[/tex] and [tex]d+e=2[/tex]
[tex]\implies d=4[/tex] and [tex]e=-2[/tex]
Rewrite [tex]2x[/tex] as [tex]+4x-2x[/tex]:
[tex]\implies x^2+4x-2x-8[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies x(x+4)-2(x+4)[/tex]
Factor out common term [tex](x+4)[/tex]:
[tex]\implies (x-2)(x+4)[/tex]
Zeros
[tex]\implies (x-2)=0\implies x=2[/tex]
[tex]\implies (x+4)=0\implies x=-4[/tex]
Vertex
[tex]\sf x-value=\dfrac{2+(-4)}{2}=-1[/tex]
[tex]\sf y-value=(-1)^2+2(-1)-8=-9[/tex]
Vertex = (-1, -9)
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Question (b)
Factor [tex]y=-x^2-9x-14[/tex]
[tex]ac=14[/tex] and [tex]d+e=-9[/tex]
[tex]\implies d=-7[/tex] and [tex]e=-2[/tex]
Rewrite [tex]-9x[/tex] as [tex]-7x-2x[/tex]:
[tex]\implies -x^2-7x-2x-14[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies -x(x+7)-2(x+7)[/tex]
Factor out common term [tex](x+7)[/tex]:
[tex]\implies (-x-2)(x+7)[/tex]
Factor [tex](-x-2):-(x+2)[/tex]
[tex]\implies -(x+2)(x+7)[/tex]
Zeros
[tex]\implies -(x+2)=0\implies -x-2=0 \implies x=-2[/tex]
[tex]\implies (x+7)=0\implies x=-7[/tex]
Vertex
[tex]\sf x-value=\dfrac{-2-7}{2}=-4.5[/tex]
[tex]\sf y-value=-(-4.5)^2-9(-4.5)-14=6.25[/tex]
Vertex = (-4.5, 6.25)
