Question

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

Answer 1

Answer:

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.

$\textsf{midpoint}=\dfrac{-4+2}{2}=-1$

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.

$\textsf{midpoint}=\dfrac{-7+(-2)}{2}=-4.5$

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)