## 17Calculus Precalculus - Intercepts of Polynomials

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The intercepts of a graph are the points where a graph crosses either the x-axis (x-intercepts) or the y-axis (y-intercepts). 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.

X-Intercepts

To determine the x-intercepts, we need to find where the graph crosses the x-axis. In these cases, the y-value is equal to zero. So in the equation, we set $$y=0$$ and then solve for the x-values. X-intercepts are also called zeros or roots. We discuss this in more detail on this page.

Y-Intercepts

To determine the y-intercepts, we need to find where the graph crosses the y-axis. In these cases, the x-value is equal to zero. So in the equation, we set $$x=0$$ and then solve for the y-values. If the equation represents a function (which all polynomials are), then there will be only one y-intercept since a function passes the vertical line test.

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 $$y = 2x + 1$$.

x-intercept $$(-1/2, 0)$$
y-intercept $$(0,1)$$

Problem Statement

Determine the x and y intercepts for $$y = 2x + 1$$.

Solution

### 2786 video

x-intercept $$(-1/2, 0)$$
y-intercept $$(0,1)$$

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$$y = x/2 - 3$$

Problem Statement

Determine the x and y intercepts for $$y = x/2 - 3$$.

x-intercept $$(6, 0)$$
y-intercept $$(0,-3)$$

Problem Statement

Determine the x and y intercepts for $$y = x/2 - 3$$.

Solution

### 2787 video

x-intercept $$(6, 0)$$
y-intercept $$(0,-3)$$

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$$2x - 3y = 6$$

Problem Statement

Determine the x and y intercepts for $$2x - 3y = 6$$.

x-intercept $$(3, 0)$$
y-intercept $$(0,-2)$$

Problem Statement

Determine the x and y intercepts for $$2x - 3y = 6$$.

Solution

### 2788 video

x-intercept $$(3, 0)$$
y-intercept $$(0,-2)$$

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$$3x + 4y = 12$$

Problem Statement

Determine the x and y intercepts for $$3x + 4y = 12$$.

x-intercept $$(4, 0)$$
y-intercept $$(0,3)$$

Problem Statement

Determine the x and y intercepts for $$3x + 4y = 12$$.

Solution

### 2789 video

x-intercept $$(4, 0)$$
y-intercept $$(0,3)$$

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$$y = x^2 + 1$$

Problem Statement

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

x-intercepts: none
y-intercepts: $$y=1$$

Problem Statement

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

Solution

To find the x-intercepts, we set $$y=0$$ and solve the the x-values.
$$\begin{array}{rcl} 0 & = & x^2+1 \\ -1 & = & x^2 \end{array}$$
Since there are no real values that solve this equation, we know that there are no real x-intercepts.

Now let's determine the y-intercepts. To do this, we set $$x=0$$ and solve.
$$y=0^2+1 = 1$$
So, the only y-intercept is $$y=1$$.

x-intercepts: none
y-intercepts: $$y=1$$

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$$y=2(x-1)(x+2)(x-3)$$

Problem Statement

Determine the intercepts (both x and y) for $$y=2(x-1)(x+2)(x-3)$$.

Solution

### 2532 video

video by PatrickJMT

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$$y=x^2-3x-18$$

Problem Statement

Determine the intercepts (both x and y) for $$y=x^2-3x-18$$.

Solution

### 2533 video

video by PatrickJMT

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$$f(x) = (x-2)(x+1)(x-4)(x+3)$$

Problem Statement

Determine the intercepts (both x and y) for $$f(x) = (x-2)(x+1)(x-4)(x+3)$$.

Solution

### 2534 video

video by MIP4U

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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

### 2535 video

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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

### 2557 video

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Really UNDERSTAND Precalculus

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

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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