Polynomial Functions

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Pre-Calculus › Polynomial Functions

Questions 1 - 10

Find the zeros and asymptotes for

$y = \frac{2x^2 + 5 x - 3 }{2x^2 + 3x - 2 }$ .

Zero: ; Asymptote:

Zeros: ; Asymptote:

Zero: $x = \frac{1}{2}$ ; Asymptotes: $x = 2, -\frac{1}{2}$

Zeros: $x = 2, \frac{1}{2}$ ; Asymptotes: $x = 3, \frac{1}{2}$

Zero: ; Asymptotes: $x = -2, \frac{1}{2}$

Explanation

To find the information we're looking for, we should factor this equation:

$y = \frac{(2 x - 1 )(x + 3 )}{(2 x - 1 )(x+ 2 )}$

This means that it simplifies to $y = \frac{x + 3 }{ x + 2 }$ .

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

$\x+3=0\x=-3$

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

$\x+2=0\x=-2$

Therefore our equation has a zero at -3 and an asymptote at -2.

Solve this equation and check your answer:

$1+\frac{n+6}{n+1}=\frac{4}{n-2}$

No solution

$n=\frac{1\pm\sqrt{73}}{2}$

$n=\frac{-7\pm\sqrt{193}}{4}$

$n=\frac{1\pm \sqrt{145}}{4}$

Explanation

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this:

$(n+1)(n-2)\cdot 1+(n+1)(n-2)\frac{n+6}{n+1}=(n+1)(n-2)\frac{4}{n-2}$

Simplify to get:

Expand to get:

Move all terms to one side and combine to get:

Use the quadratic formula to get:

$n= \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{1\pm \sqrt{1^2 - 4\cdot 2\cdot -18}}{4} = \frac{1\pm \sqrt{145}}{4}$

Find the zeros and asymptotes for

$y = \frac{2x^2 + 5 x - 3 }{2x^2 + 3x - 2 }$ .

Zero: ; Asymptote:

Zeros: ; Asymptote:

Zero: $x = \frac{1}{2}$ ; Asymptotes: $x = 2, -\frac{1}{2}$

Zeros: $x = 2, \frac{1}{2}$ ; Asymptotes: $x = 3, \frac{1}{2}$

Zero: ; Asymptotes: $x = -2, \frac{1}{2}$

Explanation

To find the information we're looking for, we should factor this equation:

$y = \frac{(2 x - 1 )(x + 3 )}{(2 x - 1 )(x+ 2 )}$

This means that it simplifies to $y = \frac{x + 3 }{ x + 2 }$ .

When the equation is in the form of a fraction, to find the zero of the function we need to set the numerator equal to zero and solve for the variable.

$\x+3=0\x=-3$

To find the asymptote of an equation with a fraction we need to set the denominator of the fraction equal to zero and solve for the variable.

$\x+2=0\x=-2$

Therefore our equation has a zero at -3 and an asymptote at -2.

Solve this equation and check your answer:

$1+\frac{n+6}{n+1}=\frac{4}{n-2}$

No solution

$n=\frac{1\pm\sqrt{73}}{2}$

$n=\frac{-7\pm\sqrt{193}}{4}$

$n=\frac{1\pm \sqrt{145}}{4}$

Explanation

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this:

$(n+1)(n-2)\cdot 1+(n+1)(n-2)\frac{n+6}{n+1}=(n+1)(n-2)\frac{4}{n-2}$

Simplify to get:

Expand to get:

Move all terms to one side and combine to get:

Use the quadratic formula to get:

$n= \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{1\pm \sqrt{1^2 - 4\cdot 2\cdot -18}}{4} = \frac{1\pm \sqrt{145}}{4}$

Factorize the following polynomial expression completely to its linear factors: $f(x)=10x^{2}-4x+15xy-6y$

$(2x+3y)\cdot (5x-2)$

$(2x+3y\cdot (5x+2)$

$(2x-3y\cdot (5x+2)$

$(2x+3)\cdot (5x-2)$

Explanation

Use the grouping method to factorize common terms: $10x^{2}-4x+15xy-6y\rightarrow 2x(5x-2)\rightarrow (2x+3y)\cdot (5x-2)$

Factorize the following polynomial expression completely to its linear factors: $f(x)=10x^{2}-4x+15xy-6y$

$(2x+3y)\cdot (5x-2)$

$(2x+3y\cdot (5x+2)$

$(2x-3y\cdot (5x+2)$

$(2x+3)\cdot (5x-2)$

Explanation

Use the grouping method to factorize common terms: $10x^{2}-4x+15xy-6y\rightarrow 2x(5x-2)\rightarrow (2x+3y)\cdot (5x-2)$

Which of the following is and accurate graph of $f(x) = 1 + \sqrt{x + 3}$ ?

Varsity2

Varsity3

Varsity4

Varsity5

Varsity6

Explanation

Remember , for .

Step 1, realize where starts: A) observe never occurs, B) zero-out the radical component of ;

$f(-3) = 1 + \sqrt{-3 + 3}$

$= 1 + \sqrt{0}$

C) The resulting point is .

Step 2, find simple points for after :

, so use ;

$f(-2) = 1 + \sqrt{-2 + 3}$

$= 1 + \sqrt{1}$

The next resulting point; .

, so use ;

$f(1) = 1 + \sqrt{1 + 3}$

$= 1 + \sqrt{4}$

The next resulting point; .

Step 3, draw a curve through the considered points.

Factorize the following polynomial expression completely to its linear factors: $f(x)=10x^{2}-4x+15xy-6y$

$(2x+3y)\cdot (5x-2)$

$(2x+3y\cdot (5x+2)$

$(2x-3y\cdot (5x+2)$

$(2x+3)\cdot (5x-2)$

Explanation

Use the grouping method to factorize common terms: $10x^{2}-4x+15xy-6y\rightarrow 2x(5x-2)\rightarrow (2x+3y)\cdot (5x-2)$

Solve this equation and check your answer:

$1+\frac{n+6}{n+1}=\frac{4}{n-2}$

No solution

$n=\frac{1\pm\sqrt{73}}{2}$

$n=\frac{-7\pm\sqrt{193}}{4}$

$n=\frac{1\pm \sqrt{145}}{4}$

Explanation

To solve this, first, find the common denominator. It is (n+1)(n-2). Multiply the entire equation by this:

$(n+1)(n-2)\cdot 1+(n+1)(n-2)\frac{n+6}{n+1}=(n+1)(n-2)\frac{4}{n-2}$

Simplify to get:

Expand to get:

Move all terms to one side and combine to get:

Use the quadratic formula to get:

$n= \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{1\pm \sqrt{1^2 - 4\cdot 2\cdot -18}}{4} = \frac{1\pm \sqrt{145}}{4}$

Which of the following is and accurate graph of $f(x) = 1 + \sqrt{x + 3}$ ?

Varsity2

Varsity3

Varsity4

Varsity5

Varsity6

Explanation

Remember , for .

Step 1, realize where starts: A) observe never occurs, B) zero-out the radical component of ;

$f(-3) = 1 + \sqrt{-3 + 3}$

$= 1 + \sqrt{0}$

C) The resulting point is .

Step 2, find simple points for after :

, so use ;

$f(-2) = 1 + \sqrt{-2 + 3}$

$= 1 + \sqrt{1}$

The next resulting point; .

, so use ;

$f(1) = 1 + \sqrt{1 + 3}$

$= 1 + \sqrt{4}$

The next resulting point; .

Step 3, draw a curve through the considered points.

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