Question

Could i have super quick help before i'm out of time?

Answer 1

Part (a)

Given quadratic:
`y=x^2+2x-8`

Factored form

To factor, find two numbers that multiply to -8 and sum to 2: 4 and -2

Rewrite the middle term of the quadratic as the sum of these number:

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

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

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

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

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

Zeros

The zeros of the quadratic polynomial are the x-coordinates of the points where the graph intersects the x-axis, i.e. when y = 0

`\implies y=0`

`\implies (x-2)(x+4)=0`

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

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

Therefore, the zeros are 2 and -4

Vertex

The x-coordinate of the vertex is the midpoint of the zeros.

`\textsf{midpoint}=(-4+2)/(2)=-1`

To find the y-coordinate of the vertex, substitute the found value of x into the given equation:

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

Therefore, the vertex is (1, -9)

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Part (b)

Given quadratic:
`y=-x^2-9x-14`

Factored form

To factor, first factor out -1:

`y=-(x^2+9x+14)`

Now find two numbers that multiply to 14 and sum to 9: 7 and 2

Rewrite the middle term of the quadratic as the sum of these number:

`y=-(x^2+2x+7x+14)`

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

`y=-(x(x+2)+7(x+2))`

Factor out the common term
`(x+2)`:

`y=-(x+7)(x+2)`

Zeros

The zeros of the quadratic polynomial are the x-coordinates of the points where the graph intersects the x-axis, i.e. when y = 0

`\implies y=0`

`\implies -(x+7)(x+2)=0`

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

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

Therefore, the zeros are -7 and -2

Vertex

The x-coordinate of the vertex is the midpoint of the zeros.

`\textsf{midpoint}=(-7+(-2))/(2)=-4.5`

To find the y-coordinate of the vertex, substitute the found value of x into the given equation:

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

Therefore, the vertex is (-4.5, 6.25)