19. Properties of Graphs

b. Value & Limit Information

1. Intercepts & Positive and Negative Values

We start by identifying where the function is positive, negative or zero and where it crosses the \(y\)-axis.

An \(x\)-intercept of \(f(x)\) is a point where the graph crosses the \(x\)-axis.
A \(y\)-intercept of \(f(x)\) is a point where the graph crosses the \(y\)-axis.

def_intercepts

To find the \(x\)-intercepts, we solve the equation \(f(x)=0\). There can be zero, one or many \(x\)-intercepts.

To find the \(y\)-intercepts, we simply plug \(x=0\) into the function, i.e. the \(y\)-intercept is \(f(0)\) if it is defined. There is at most one \(y\)-intercept.

Consider the function \(f(x)=2x-\dfrac{16}{x-2}\). First find any points where the function is undefined. Then find the \(x\)-intercepts, the \(y\)-intercept and the intervals where \(f(x)\) is positive or negative.

We first notice that the function is undefined at \(x=2\). Next, to find the \(x\)-intercepts, we rewrite the function as: \[ f(x)=\dfrac{2x(x-2)-16}{x-2}=\dfrac{2x^2-4x-16}{x-2}=\dfrac{2(x+2)(x-4)}{x-2} \] A fraction can only be \(0\) if the numerator is \(0\). So, we solve \(2(x+2)(x-4)=0\). So the \(x\)-intercepts are \(x=-2\) and \(x=4\).
To find the \(y\)-intercepts, we plug in \(x=0\): \[ f(0)=\dfrac{2(0+2)(0-4)}{0-2}=8 \] To determine where \(f(x)\) is positive or negative, we first plot the interesting points on a number line. These are the \(x\)-intercepts, \(x=-2\) and \(x=4\), and the places where function is undefined, \(x=2\): eg_2x-16_(x-2)_number_line_crit Since \(f\) is continuous except where it is undefined, its sign cannot change without passing through \(0\) or a point where it is undefined. So there must be a single sign on each of the intervals \((-\infty,-2)\), \((-2,2)\), \((2,4)\) and \((4,\infty)\). We can find the signs by plugging in a number in each interval: \[\begin{aligned} f(-3)&=\dfrac{2(-3+2)(-3-4)}{-3-2}=-\,\dfrac{14}{5} \quad & f(0)&=\dfrac{2(0+2)(0-4)}{0-2}=8 \\ f(3)&=\dfrac{2(3+2)(3-4)}{3-2}=-10 & f(5)&=\dfrac{2(5+2)(5-4)}{5-2}=\dfrac{14}{3} \end{aligned}\] So \(f\) is positive on the intervals \((-2,2)\) and \((4,\infty)\). And \(f\) is negative on the intervals \((-\infty,-2)\) and \((2,4)\). We add these facts to the number line: eg_2x-16_(x-2)_number_line_signs

We will eventually have enough information to plot the function. But so far we can see the \(x\)-intercepts at \(x=-2\) and \(x=4\), the \(y\)-intercept at \(y=8\) and the function is positive on \((-2,2)\) and \((4,\infty)\) and negative on \((-\infty,-2)\) and \((2,4)\). It remains to examine the behavior at \(x=\pm\infty\) and \(y=\pm\infty\), to show the function is everywhere increasing, concave up on the left, concave down on the right, and has no local maxima or minima.

eg_2x-16_(x-2)

Consider the function \(f(x)=1+\dfrac{4}{x^2}\). First find any points where the function is undefined. Then find the \(x\)-intercepts, the \(y\)-intercept and the intervals where \(f(x)\) is positive or negative. Finally, draw a number line which shows the \(x\)-intercepts (marked by a \(\bullet\)), the points where the function is undefined (marked by an \(\times\)) and the signs of the function between these points.

The function is undefined at \(x=0\). There are no \(x\)-intercepts nor \(y\)-intercept. The function is always positive. ex_1+4_x^2_number_line_signs

The function is undefined at \(x=0\). So there is no \(y\)-intercept. The function is always positive. So there are no \(x\)-intercepts. We plot these facts on a number line: ex_1+4_x^2_number_line_signs

We will eventually have enough information to plot the function. But so far we can see there no intercepts and the function is everywhere positive.

ex_1+4_x^2

Consider the function \(f(x)=\dfrac{2x^2-10x+12}{x^2-4}\). First find any points where the function is undefined. Then find the \(x\)-intercepts, the \(y\)-intercept and the intervals where \(f(x)\) is positive or negative. Finally, draw a number line which shows the \(x\)-intercepts (marked by a \(\bullet\)), the points where the function is undefined (marked by an \(\times\) if it is \(\pm\infty\) and an \(\circ\) if it is a hole) and the signs of the function between these points.

The function is undefined at \(x=\pm2\). \(x=2\) is a hole while at \(x=-2\) the function approaches \(\pm\infty\). The \(x\)-intercept is at \(x=3\). The \(y\)-intercept is \(f(0)=-3\). The function is positive on \((-\infty,-2)\), negative on \((-2,3)\) and positive again on \((3,\infty)\). ex_2x^2-10x+12_x^2-4_number_line_signs

We first factor the numerator and denominator and cancel any common factors: \[ f(x)=\dfrac{2x^2-10x+12}{x^2-4}=\dfrac{2(x-2)(x-3)}{(x-2)(x+2)} =\dfrac{2(x-3)}{x+2} \] So the function is undefined at \(x=\pm2\) because the original denominator is \(0\). However, \(x=2\) is a hole because it cancels out and it approaches the finite number: \[ \lim_{x\to2} f(x)=\lim_{x\to2} \dfrac{2(x-3)}{x+2}=-\,\dfrac{1}{2} \] On the other hand, \(x=-2\) is a place where the function approaches \(\pm\infty\) since there is a \(0\) in the denominator even after factoring. (It is a vertical asymptote which we study on the next page.) The numerator is \(0\) when \(x=3\) which is the \(x\)-intercept. The \(y\)-intercept is \(f(0)=\dfrac{2(-3)}{2}=-3\). We plot these on a number line: ex_2x^2-10x+12_x^2-4_number_line_crit To find where the function is positive or negative, we evaluate if at a point in each interval: \[\begin{aligned} f(-4)&=\dfrac{2(-4-3)}{-4+2}=7 \quad & f(0)&=\dfrac{2(-3)}{2}=-3 \\ f(4)&=\dfrac{2(4-3)}{4+2}=\dfrac{1}{3} \end{aligned}\] So the function is positive on \((-\infty,-2)\), negative on \((-2,3)\) and positive again on \((3,\infty)\). ex_2x^2-10x+12_x^2-4_number_line_signs

We will eventually have enough information to plot the function. But so far we can see the \(x\)-intercept at \(x=3\), the \(y\)-intercept at \(y=-3\) and that the function is positive on \((-\infty,-2)\) and \((3,\infty)\) and negative on \((-2,3)\).

ex_2x^2-10x+12_x^2-4

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