How do you find the x and y intercepts of x^2+x-2/x^2-3x-4?
1 answer:
First equal the equation to f(x) or y
f(x)= (x²+x-2)÷(x²-3x-4)
To find y-intercept, x coordinates have to be 0
f(0)=(0²+0-2)÷(0²-3(0)-4)
f(0)=-2/-4
f(0)=1/2 (y-intercept)
To find x-intercept, make y=0
0=(x²+x-2)÷(x²-3x-4)
0((x²-3x-4)=(x²+x-2)
0=x²+x-2
0=(x-1)(x+2)
x-1
x-1=0
x=1
x+2=0
x=-2
The x-intercepts would be 1, -2
Hope I helped :)
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