Question

could i have some super quick quick help? graph part isn't necessary

Answer 1

y=4x²-1

#1

• y=(2x)²-1²
• y=(2x+1)(2x-1)

#2

Zeros are 1/2,-1/2

#3

Convert to vertex form y=a(x-h)²+k

• y=(4x+0)²-1

Vertex(h,k)=(0,-1)

#4

Graph attached

Answer 2

Answer:

i)  $y=(2x+1)(2x-1)$

ii)  $x=-\dfrac12, x=\dfrac12$

iii)   (0, -1)

iv)  see attached

Step-by-step explanation:

Given quadratic:  $y=4x^2-1$

Factored form

We can use the Difference of Two Squares to factor the given quadratic.

$a^2-b^2=(a+b)(a-b)$

Therefore,

$a^2=4x^2=(2x)^2\implies a=2x$

$b^2=1=1^2 \implies b=1$

So the relation in factored form is:

$y=(2x+1)(2x-1)$

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

$\implies y=0$

$\implies (2x+1)(2x-1)=0$

$\implies (2x+1)=0\implies x=-\dfrac12$

$\implies (2x-1)=0\implies x=\dfrac12$

Therefore, the zeros are -1/2 and 1/2

Vertex

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

$\textsf{midpoint}=\dfrac{-\frac12+\frac12}{2}=0$

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

$\implies y=4(0)^2-1=-1$

Therefore, the vertex is (0, -1)