1.1 X and y-intercepts

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Finding x&y intercepts

A rational function is a function of the form $R(x)=\frac{p(x)}{q(x)}$ , where p and q are polynomial functions and $q\neq 0$ .

The domain is all real numbers except for numbers that make the denominator = 0.

x-intercepts are the points at which the graph crosses the x-axis. They are also known as roots, zeros, or solutions.

To find x-intercepts, let y (or f(x)) = 0 and solve for x. In rational functions, this means that you are multiplying by 0 so to find the x-intercept, just set the numerator (the top of the fraction) equal to 0 and solve for x.

Remember: x-intercepts are points that look like (x,0)

For $y=\frac{x-1}{x-2}$ find the x-intercept

The x-intercept is (1,0) since $x-1=0$ , $x=1$

The y-intercept is the point where the graph crosses the y-axis. If the graph is a function, there is only one y-intercept (and it only has ONE name)

To find the y-intercept (this is easier than the x-intercept), let x = 0. Plug in 0 for x in the equation and simplify.

Remember: y-intercepts are points that look like (0,y)

For $y=\frac{x+1}{x-2}$ find the y-intercept

The y-intercept is (0, $\frac{-1}{2}$ ) since $\frac{0+1}{0-2}=\frac{-1}{2}$

Find the x- and y-intercepts of the following:

$y=\frac{1}{x+2}$

x-intercept: None since $1\neq 0$

y-intercept: (0, $\frac{1}{2}$ ) since $\frac{1}{0+2}=\frac{1}{2}$

$y=\frac{1-3x}{1-x}$

x-intercept: ( $\frac{1}{3}$ ,0) since $1-3x=0$ , $-3x=-1$ , $x=\frac{1}{3}$

y-intercept: (0,1) since $\frac{1-3\times 0}{1-0}=1$

$y=\frac{x^{2}}{x^{2}+9}$

x-intercept: (0,0) since $x^{2}=0$ , $x=0$

y-intercept: (0,0) since $\frac{0^{2}}{0^{2}+9}=\frac{0}{9}=0$ or because the x-intercept is (0,0)

$y=\frac{\sqrt{x+1}}{(x-2)^{2}}$

x-intercept: (-1,0) since $\sqrt{x+1}=0$ , $x+1=0$ , $x=-1$

y-intercept: (0, $\frac{1}{4}$ ) since $\frac{\sqrt{0+1}}{(0-2)^{2}}=\frac{\sqrt{1}}{-2^{2}}=\frac{1}{4}$

$y=\frac{3x}{x^{2}-x-2}$

x-intercept: (0,0) since $3x=0$ , $x=0$

y-intercept: (0,0) since the x-intercept is (0,0)

$y=\frac{1}{x-3}+1$

x-intercept: (2,0) since $\frac{1}{x-3}+1=0$ , $\frac{1}{x-3}=-1$ , $-x+3=1$ , $-x=-2$ , $x=2$

y-intercept: (0, $\frac{2}{3}$ ) since $\frac{1}{0-3}+1=\frac{-1}{3}+1=\frac{2}{3}$

$y=\frac{x^{2}-4}{\sqrt{x+1}}$

x-intercepts: (-2,0), (2,0) since $x^{2}-4=0$ , $x^{2}=4$ , $x=-2$ , $x=2$

y-intercept: (0, -4) since $\frac{0^{2}-4}{\sqrt{0+1}}=\frac{-4}{\sqrt{1}}=-4$

$y=4+\frac{5}{x^{2}+2}$

x-intercept: None since $4+\frac{5}{x^{2}+2}=0$ , $\frac{5}{x^{2}+2}=-4$ , $-4x^{2}-8=5$ , $-4x^{2}=13$ , $x^{2}=\frac{13}{4}$ , a number squared will never be a negative number, so there is no x-intercept

y-intercept: (0, $\frac{13}{2}$ ) since $4+\frac{5}{0^{2}+2}=4+\frac{5}{2}=\frac{13}{2}$

$y=\frac{\sqrt{5x-2}}{x-3}$

x-intercept: ( $\frac{2}{5}$ , 0) since $\sqrt{5x-2}=0$ , $5x-2=0$ , $5x=2$ , $x=\frac{2}{5}$

y-intercept: None since $y=\frac{\sqrt{5\times 0-2}}{0-3}$ takes the square root of a negative number.

$y=\frac{x^{3}-8}{x^{2}+1}$

x-intercept: (2,0) since $x^{3}-8=0$ , $(x-2)(x^{2}+2x+4)=0$ , $x=2$

y-intercept: (0,-8) since $\frac{0^{3}-8}{0^{2}+1}=\frac{-8}{1}=-8$

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