Question

Consider the function f(x) = the quantity x squared minus x minus 12 all over the quantity x plus 3. Describe the graph of this function. Include all discontinuities, intercepts, and the basic shape of the graph.

Answer 1

First, substitute the value of x into the function, that would be that is now f(x) = x-4, cancelling -3. We have to do this to get the point of discontinuity

Your remaining function now is f(x) = x-4, which is a line. Find the value of the x-intercept by making y to 0 and vice versa when finding the value of the y-intercept.

 

After cancelling the (x+3) from both the numerator and the denominator, your point discontinuity happens when the factor is equal to zero. This happens after you substitute x= -3 into the new function that is y=x-4, so you will get -7. So the point discontinuity lies at the f(-3)=-7 or (-3, -7).

Find the value of the x- intercept and y-intercept, you equate it with zero, that would be

0= x-4 x=4 (4, 0).

 y= 0 - 4 y= -4 (0, -4).

Answer 2

Answer:

The graph of f(x) is discontinuous at x= -3. The x and y-intercepts or he function f(x) are (4,0) and (0,-4) respectively. The graph of the function is a straight line.

Step-by-step explanation:

The given function is

$f(x)=\frac{x^2-x-12}{x+3}$

$f(x)=\frac{x^2-4x+3x-12}{x+3}$

$f(x)=\frac{x(x-4)+3(x-4)}{x+3}$

$f(x)=\frac{(x-4)(x+3)}{x+3}$

Cancel out the common factor (x+3).

$f(x)=x-4$

It is a linear equation and it will give a straight line.

Put x=0, to find the y-intercept.

$f(x)=0-4$

$f(x)=-4$

Therefore the y-intercept is (0,-4).

Put x=0, to find the x-intercept.

$0=x-4$

$4=x$

Therefore the x-intercept is (4,0).

At x= -3 the value of denominator is 0, therefore the function is undefined for x= -3.

As x approaches toward x=-3 from left and right, the value of function approaches towards y=-7.

The function is discontinuous at x= -3.