<em><u>Question:</u></em>
Consider the following equation: f(x)=x^2+4\4x^2-4x-8 name the vertical asymptote(s)
<em><u>Answer:</u></em>
The vertical asymptotes are x = 2 or x = -1
<em><u>Solution:</u></em>
Given that,
For vertical asymptotes, set the denominator to zero and then solve the quadratic equation
Thus the vertical asymptotes are x = - 1 and x = 2
Answer:
idk its hard to me too
Step-by-step explanation:
He still has 83 miles to travel
<span><span><span>x3</span>+3x5</span>=x5</span><span>3x5+x3=x5</span>Subtract x^5 from both sides.<span>3x5+x3−<span>x5</span>=x5−<span>x5</span></span><span>2x5+x3=0</span>Factor left side of equation.<span><span><span>x3</span>(2x2+1)</span>=0</span>Set factors equal to 0.<span><span><span>x3</span>=0 or 2x2+1</span>=0</span><span>x=<span>0</span></span>
26 = 3x - 2 - 7x
26 + 2 = -4x
28 / -4 = x
x = -7
Best answer me please!
