pre calc homework

Question

f(x)= 2x^2+7x-4/3x^2+4x-15

 A y intercept at the point    

2) x intercepts at the point(s)    

3) Vertical asymptotes at x =    

4) Horizontal asymptote at  y = 

Answer 1

Factor the numerator and denominator with prefered method:

f(x) = (2x^2+7x-4)/(3x^2+4x-15)

f(x) = [(2x-1)(x+4)]/[(3x-5)(x+3)]

The x-intercepts are where f(x)=0, so we can just solve 0=(2x-1)(x+4) and ignore the denominator part. Thus, the x-intercepts are x=1/2 and x=-4, or (1/2,0) and (-4,0)

The y-intercept is where x=0, so f(0) = [(2(0)-1)(0+4)]/[(3(0)-5)(0+3)] = [(-1)(4)]/[(-5)(3)] = -4/-15 = 4/15, which means the y-intercept is y=4/15, or (0,4/15)

Vertical asymptotes are where 3x-5=0 and x+3=0, so these are x=5/3 and x=-3

The horizontal asymptote will be the quotient of the leading coefficients since the degrees of the numerator and denominator are the same, so this would be y=2/3.

Hope this helped!

Answer 2

Hi Ashley

There are some rules associated with finding intercepts and calculating asymptotes.

Take a good look at your function, it contains a quadratic in the numerator and one in the denominator, can they be factored, this will be important for your answers. It would also be a good idea to label your quadratics according to Standard (or General) form. Finally graph your function to see it and confirm the information below

f(x)= ax² + bx + c

For your function

f(x) = (2x²+ 7x -4)/(3x² + 4x -15)

The y intercept can be found by setting x equal to zero

f(x) = (2*(0)² + 7*(0) - 4)/(3*(0)² + 4*(0) - 15) = -4/-15

y- intercept is (0, 4/15)

To find the x intercepts you set y or f(x) equal to zero

0 = (2x²+ 7x -4)/(3x² + 4x -15)

Multiply both sides by the denominator to give

0 = 2x² + 7x - 4

Can this quadratic be factored

(2x -1)(x + 4) = 0

x = 1/2

x = -4

x intercepts are (1/2,0) (-4, 0)

For the vertical asymptotes the denominator cannot be zero, going back to your original function

f(x) = (2x²+ 7x -4)/(3x² + 4x -15)

3x² + 4x -15 cannot equal zero

We factored the quadratic in the numerator, what about this one in the denominator

3x² + 4x -15

(3x - 5)(x + 3)

So f(x) can be expressed in factored form below

f(x) = (2x -1)(x+4)/(3x -5)(x +3)

To find the vertical asymptotes we set the denominator equal to zero, for calculation purposes

(3x -5)(x + 3)=0

Solving the denominator for x,

x = -3 and x 5/3 would make the denominator zero and we can’t have division by zero, so

Vertical Asymptotes would occur at x = -3 and x = 5/3

For the horizontal asymptotes go back to the rules, notice the exponent in the numerator and denominator,

f(x) = (2x²+ 7x -4)/(3x² + 4x -15)

The exponent in the numerator, n is 2 and the exponent in the denominator d, is also 2, check the rules for when the exponents are the same to determine the horizontal asymptotes. When n=d the rule says the horizontal asymptotes y =a_n/b_d this goes back to using the leading coefficients in your quadratic equation. The leading coefficient in the numerator is represented by a_n and leading coefficient in the denominator is represented by b_d.

a_n = 2, b_d = 3 (updated)

Can you continue from here?

Of course you should graph your function at Desmos.com or on a Graphing calculator to confirm the information above,