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

Could i have some super quick quick help? graph part isn't necessary-example-1
User AZhao
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2 Answers

3 votes

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

Could i have some super quick quick help? graph part isn't necessary-example-1
User Jkitchen
by
8.3k points
10 votes

Answer:

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

ii)
x=-\frac12, x=\frac12

iii) (0, -1)

iv) see attached

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=-\frac12


\implies (2x-1)=0\implies x=\frac12

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

Vertex

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


\textsf{midpoint}=(-\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)

Could i have some super quick quick help? graph part isn't necessary-example-1
User Uzzar
by
8.8k points

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