Algebra - Identify Zeros, Factor and Graph Polynomials: CCSS.Math.Content.HSA-APR.B.3

What are the -intercept(s) of the function?

$\text{x-intercepts: }3,4$

$\text{x-intercepts: }-3,4$

$\text{x-intercepts: }3,-4$

$\text{x-intercepts: }-3,-4$

$\text{x-intercepts: }6,2$

Explanation

To find the -intercept of a function, first recall that the -intercept represents the points where the graph of the function crosses the -axis. In other words where the function has a value equal to zero.

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

and are factors of and when added together results in .

For the given function,

the coefficients are,

$\b=-7 \c=12$

therefore the factors of that have a sum of are,

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\x-3=0\rightarrow x=3 \x-4=0\rightarrow x=4$

To verify, graph the function.

Screen shot 2016 03 08 at 12.13.08 pm

The graph crosses the -axis at 3 and 4, thus verifying the results found by factorization.

What are the -intercept(s) of the function?

$\text{x-intercepts:}-6,-3$

$\text{x-intercepts:}-6,3$

$\text{x-intercepts: }6,-3$

$\text{x-intercepts: }6,3$

$\text{x-intercepts: }2,9$

Explanation

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

and are factors of and when added together results in .

For the given function,

the coefficients are,

$\b=9 \c=18$

therefore the factors of that have a sum of are,

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\x+3=0\rightarrow x=-3 \x+6=0\rightarrow x=-6$

To verify, graph the function.

Screen shot 2016 03 08 at 12.27.49 pm

The graph crosses the -axis at -3 and -6, thus verifying the results found by factorization.

What are the -intercept(s) of the function?

$\text{x-intercepts: }-4,2$

$\text{x-intercepts: }4,2$

$\text{x-intercepts: }-4,-2$

$\text{x-intercepts: }4,-2$

$\text{x-intercepts: }-1,8$

Explanation

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

and are factors of and when added together results in .

For the given function,

the coefficients are,

$\b=2 \c=-8$

therefore the factors of that have a sum of are,

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\x-2=0\rightarrow x=2 \x+4=0\rightarrow x=-4$

To verify, graph the function.

Screen shot 2016 03 08 at 12.52.32 pm

The graph crosses the -axis at -4 and 2, thus verifying the results found by factorization.

What are the -intercept(s) of the function?

$\text{x-intercepts: }-6,-1$

$\text{x-intercepts: }-6,1$

$\text{x-intercepts: }6,-1$

$\text{x-intercepts: }6,1$

$\text{x-intercepts: }3,2$

Explanation

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

and are factors of and when added together results in .

For the given function,

the coefficients are,

$\b=7 \c=6$

therefore the factors of that have a sum of are,

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\x+1=0\rightarrow x=-1 \x+6=0\rightarrow x=-6$

To verify, graph the function.

Screen shot 2016 03 08 at 1.27.02 pm

The graph crosses the -axis at -6 and -1, thus verifying the results found by factorization.

What are the -intercept(s) of the function?

$\text{x-intercept: }1$

$\text{x-intercept: }-1$

$\text{x-intercepts: }1,-1$

$\text{x-intercept: }0$

$\text{x-intercepts: }0,1$

Explanation

One technique that can be used is factorization. In general form,

$y=x^2+bx+c\rightarrow(x+c_1)(x+c_2)$

where,

and are factors of and when added together results in .

For the given function,

the coefficients are,

$\b=-2 \c=1$

therefore the factors of that have a sum of are,

Now find the -intercepts of the function by setting each binomial equal to zero and solving for .

$\x-1=0\rightarrow x=1$

To verify, graph the function.

Screen shot 2016 03 09 at 9.54.14 am

The graph crosses the -axis at 1, thus verifying the result found by factorization.

Find the zeros of $- 7 x^{2} + 34 x + 16$

$\frac{17}{7} + \frac{\sqrt{401}}{7} , - \frac{\sqrt{401}}{7} + \frac{17}{7}$

$2 \sqrt{401} , - 2 \sqrt{401}$

There are no real roots

Explanation

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$

In this case , , and

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * -7 * 16 }}{ 2 * -7 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - -448 }}{ -14 }$

$x =\frac{ -34 \pm \sqrt{ 1604 }}{ -14 }$

$x =\frac{ -34 \pm 2 \sqrt{401} }{ -14 }$

Now we split this up into two equations.

$x =\frac{ -34 + 2 \sqrt{401} }{ -14 }$

$x =\frac{ -34 - 2 \sqrt{401} }{ -14 }$

So our zeros are at

$x = \frac{17}{7} + \frac{\sqrt{401}}{7}$

$x = - \frac{\sqrt{401}}{7} + \frac{17}{7}$

Find the zeros of $4 x^{2} + 29 x + 19$

$- \frac{29}{8} - \frac{\sqrt{537}}{8} , - \frac{29}{8} + \frac{\sqrt{537}}{8}$

$\sqrt{537} , - \sqrt{537}$

There are no real roots

Explanation

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$

In this case , , and

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -29 \pm \sqrt{ 29 ^2 - 4 * 4 * 19 }}{ 2 * 4 }$

$x =\frac{ -29 \pm \sqrt{ 841 - 304 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{ 537 }}{ 8 }$

$x =\frac{ -29 \pm \sqrt{537} }{ 8 }$

Now we split this up into two equations.

$x =\frac{ -29 + \sqrt{537} }{ 8 }$

$x =\frac{ -29 - \sqrt{537} }{ 8 }$

So our zeros are at

$x = - \frac{29}{8} - \frac{\sqrt{537}}{8}$

$x = - \frac{29}{8} + \frac{\sqrt{537}}{8}$

Find the zeros of $- 5 x^{2} + 30 x - 16$

$- \frac{\sqrt{145}}{5} + 3 , \frac{\sqrt{145}}{5} + 3$

$2 \sqrt{145} , - 2 \sqrt{145}$

There are no real roots

Explanation

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$

In this case , and

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -30 \pm \sqrt{ 30 ^2 - 4 * -5 * -16 }}{ 2 * -5 }$

$x =\frac{ -30 \pm \sqrt{ 900 - 320 }}{ -10 }$

$x =\frac{ -30 \pm \sqrt{ 580 }}{ -10 }$

$x =\frac{ -30 \pm 2 \sqrt{145} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -30 + 2 \sqrt{145} }{ -10 }$

$x =\frac{ -30 - 2 \sqrt{145} }{ -10 }$

So our zeros are at

$x = - \frac{\sqrt{145}}{5} + 3$

$x = \frac{\sqrt{145}}{5} + 3$

Find the zeros of $9 x^{2} + 15 x - 52$

$- \frac{5}{6} + \frac{\sqrt{233}}{6} , - \frac{\sqrt{233}}{6} - \frac{5}{6}$

$3 \sqrt{233} , - 3 \sqrt{233}$

There are no real roots

Explanation

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation

$a x^{2} + b x + c = 0$

In this case , and

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -15 \pm \sqrt{ 15 ^2 - 4 * 9 * -52 }}{ 2 * 9 }$

$x =\frac{ -15 \pm \sqrt{ 225 - -1872 }}{ 18 }$

$x =\frac{ -15 \pm \sqrt{ 2097 }}{ 18 }$

$x =\frac{ -15 \pm 3 \sqrt{233} }{ 18 }$

Now we split this up into two equations.

$x =\frac{ -15 + 3 \sqrt{233} }{ 18 }$

$x =\frac{ -15 - 3 \sqrt{233} }{ 18 }$

So our zeros are at

$x = - \frac{5}{6} + \frac{\sqrt{233}}{6}$

$x = - \frac{\sqrt{233}}{6} - \frac{5}{6}$

Find the zeros of $x^{2} + 35 x - 67$

$\frac{35}{2} + \frac{\sqrt{1493}}{2} , - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

$\sqrt{1493} , - \sqrt{1493}$

There are no real roots

Explanation

In order to find the zeros, we can use the quadratic formula.

Recall the quadratic formula.

$x =\frac{ - b \pm \sqrt{- 4 a c + b^{2}} }{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ , correspond to the coefficients in the equation.

$a x^{2} + b x + c = 0$

In this case , and

We plug in these values into the quadratic formula, and evaluate them.

$x =\frac{ -35 \pm \sqrt{ 35 ^2 - 4 * 1 * -67 }}{ 2 * 1 }$

$x =\frac{ -35 \pm \sqrt{ 1225 - -268 }}{ 2 }$

$x =\frac{ -35 \pm \sqrt{ 1493 }}{ 2 }$

$x =\frac{ -35 \pm \sqrt{1493} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -35 + \sqrt{1493} }{ 2 }$

$x =\frac{ -35 - \sqrt{1493} }{ 2 }$

So our zeros are at

$x = - \frac{35}{2} + \frac{\sqrt{1493}}{2}$

$x = - \frac{\sqrt{1493}}{2} - \frac{35}{2}$

Identify Zeros, Factor and Graph Polynomials: CCSS.Math.Content.HSA-APR.B.3

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