The intercepts of a graph are the points where a graph crosses either the xaxis (xintercepts) or the yaxis (yintercepts). In calculus, we always work with real numbers unless explictly stated otherwise. So this page discusses only real x and y intercepts of polynomials. For intercepts of rational functions, see this page.
XIntercepts
To determine the xintercepts, we need to find where the graph crosses the xaxis. In these cases, the yvalue is equal to zero. So in the equation, we set \(y=0\) and then solve for the xvalues. Xintercepts are also called zeros or roots. We discuss this in more detail on this page.
YIntercepts
To determine the yintercepts, we need to find where the graph crosses the yaxis. In these cases, the xvalue is equal to zero. So in the equation, we set \(x=0\) and then solve for the yvalues. If the equation represents a function (which all polynomials are), then there will be only one yintercept since a function passes the vertical line test.
Checking Your Answer
It is always a good idea to check your answer. Doing so may result in an increase of your grade by one full letter grade. For these problems, plotting is a good way to check your answer.
Practice
Unless other instructed, determine the intercepts (both x and y) for these functions.
\( y = 2x + 1 \)
Problem Statement 

Determine the x and y intercepts for \( y = 2x + 1 \).
Final Answer 

xintercept \((1/2, 0)\)
yintercept \((0,1)\)
Problem Statement 

Determine the x and y intercepts for \( y = 2x + 1 \).
Solution 

Final Answer 

xintercept \((1/2, 0)\) 
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\( y = x/2  3 \)
Problem Statement 

Determine the x and y intercepts for \( y = x/2  3 \).
Final Answer 

xintercept \((6, 0)\)
yintercept \((0,3)\)
Problem Statement 

Determine the x and y intercepts for \( y = x/2  3 \).
Solution 

Final Answer 

xintercept \((6, 0)\) 
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\( 2x  3y = 6 \)
Problem Statement 

Determine the x and y intercepts for \( 2x  3y = 6 \).
Final Answer 

xintercept \((3, 0)\)
yintercept \((0,2)\)
Problem Statement 

Determine the x and y intercepts for \( 2x  3y = 6 \).
Solution 

Final Answer 

xintercept \((3, 0)\) 
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\( 3x + 4y = 12 \)
Problem Statement 

Determine the x and y intercepts for \( 3x + 4y = 12 \).
Final Answer 

xintercept \((4, 0)\)
yintercept \((0,3)\)
Problem Statement 

Determine the x and y intercepts for \( 3x + 4y = 12 \).
Solution 

Final Answer 

xintercept \((4, 0)\) 
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\( y = x^2 + 1 \)
Problem Statement 

Determine the intercepts (both x and y) for \( y = x^2 + 1 \).
Final Answer 

xintercepts: none
yintercepts: \(y=1\)
Problem Statement 

Determine the intercepts (both x and y) for \( y = x^2 + 1 \).
Solution 

To find the xintercepts, we set \(y=0\) and solve the the xvalues.
\(\begin{array}{rcl} 0 & = & x^2+1 \\ 1 & = & x^2 \end{array} \)
Since there are no real values that solve this equation, we know that there are no real xintercepts.
Now let's determine the yintercepts. To do this, we set \(x=0\) and solve.
\( y=0^2+1 = 1\)
So, the only yintercept is \(y=1\).
Final Answer 

xintercepts: none 
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\( y=2(x1)(x+2)(x3) \)
Problem Statement 

Determine the intercepts (both x and y) for \( y=2(x1)(x+2)(x3) \).
Solution 

video by PatrickJMT 

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\( y=x^23x18 \)
Problem Statement 

Determine the intercepts (both x and y) for \( y=x^23x18 \).
Solution 

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\( f(x) = (x2)(x+1)(x4)(x+3) \)
Problem Statement 

Determine the intercepts (both x and y) for \( f(x) = (x2)(x+1)(x4)(x+3) \).
Solution 

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\( f(x) = 2x^3 + 2x^2  24x \)
Problem Statement 

Determine the intercepts (both x and y) for \( f(x) = 2x^3 + 2x^2  24x \).
Solution 

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\( y = x^2 + 4x  4 \)
Problem Statement 

Determine the intercepts (both x and y) for \( y = x^2 + 4x  4 \).
Solution 

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Really UNDERSTAND Precalculus
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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