What are the zeros of the function f(c)=x^2+2x-8/x^2-2x-8

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aspenwheeler6804

03.06.2023

Mathematics

Answer with explanation:

The given function : f(x)=x^2+2x-3

Using completing the squares, we have

f(x)=x^2+2x+1-1-3 [∵ (x+1)^2=x^2+2x+1 ]

f(x)=(x+1)^2-4 (1)

Comparing (1) to the standard vertex form f(x)=(x-h)^2+k , the vertex of function is at (h,k)=(-1,-4)

For x-intercept, put f(x)=0 in (1), we get

$0=(x+1)^2-4\\\\\Rightarrow\ (x+1)^2=4$

Square root on both sides, we get

$x+1=\pm2\\\\x+1=-2\ or\ \ x+1=2\\\\=x=-3\ \ or\ x=1$

∴ x intercepts : x= (-3,0) and (1,0)

For y-intercept put x=0 in (1), we get

f(1)=(1)^2-4=1-4=-3

∴ y intercept : (0,-3)

Axis of symmetry : $\dfrac{-b}{2a}$

In f(x)=x^2+2x-3 , a=1 and b=2

Axis of symmetry= $\dfrac{-2}{2(1)}=-1$

Graph f(x) = x2 + 2x - 3, label the function’s x-intercepts, y-intercept and vertex with their coord

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