Question

i don't have much time left could someone help me super quickly (trinomials) btw you could get extra 50 points from a previous question on my profile (only if you want) i would really appreciate if i could have help understanding things

Answer 1

(a)
`(x-2)(x+4)`

zeros: x = 2, x = -4

vertex: (-1, -9)

(b)
`-(x+2)(x+7)`

zeros: x = -2, x = -7

vertex = (-4.5, 6.25)

Explanation:

To factor a quadratic in the form
`ax^2+bx+c`:

• Find 2 two numbers (d and e) that multiply to
  `ac` and sum to
  `b`
• Rewrite
  `b` as the sum of these 2 numbers:
  `d+e=b`
• 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
`(ax+b)(cx+d)`

• Set each of the parentheses to zero and solve for
  `x`

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
`y=x^2+2x-8`

`ac=-8` and
`d+e=2`

`\implies d=4` and
`e=-2`

Rewrite
`2x` as
`+4x-2x`:

`\implies x^2+4x-2x-8`

Factorize the first two terms and the last two terms separately:

`\implies x(x+4)-2(x+4)`

Factor out common term
`(x+4)`:

`\implies (x-2)(x+4)`

Zeros

`\implies (x-2)=0\implies x=2`

`\implies (x+4)=0\implies x=-4`

Vertex

`\sf x-value=(2+(-4))/(2)=-1`

`\sf y-value=(-1)^2+2(-1)-8=-9`

Vertex = (-1, -9)

-------------------------------------------------------------------------------------

Question (b)

Factor
`y=-x^2-9x-14`

`ac=14` and
`d+e=-9`

`\implies d=-7` and
`e=-2`

Rewrite
`-9x` as
`-7x-2x`:

`\implies -x^2-7x-2x-14`

Factorize the first two terms and the last two terms separately:

`\implies -x(x+7)-2(x+7)`

Factor out common term
`(x+7)`:

`\implies (-x-2)(x+7)`

Factor
`(-x-2):-(x+2)`

`\implies -(x+2)(x+7)`

Zeros

`\implies -(x+2)=0\implies -x-2=0 \implies x=-2`

`\implies (x+7)=0\implies x=-7`

Vertex

`\sf x-value=(-2-7)/(2)=-4.5`

`\sf y-value=-(-4.5)^2-9(-4.5)-14=6.25`

Vertex = (-4.5, 6.25)