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

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Question

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

Answer

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,

$\b=-1 \c=-6$

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 .

$\x-3=0\rightarrow x=3 \x+2=0\rightarrow x=-2$

To verify, graph the function.

The graph crosses the -axis at -2 and 3, thus verifying the results found by factorization.

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Question

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

Answer

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=9 \c=18$

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=2 \c=-8$

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=2 \c=1$

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=7 \c=6$

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=4 \c=3$

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 .

$\x+1=0\rightarrow x=-1 \x+3=0\rightarrow x=-3$

To verify, graph the function.

The graph crosses the -axis at -1 and -3, thus verifying the results found by factorization.

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=-2 \c=1$

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.

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

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=-8 \c=7$

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 .

$\x-1=0\rightarrow x=1 \x-7=0\rightarrow x=7$

To verify, graph the function.

The graph crosses the -axis at 1 and 7, thus verifying the results found by factorization.

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=-5 \c=6$

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 .

$\x-2=0\rightarrow x=2 \x-3=0\rightarrow x=3$

To verify, graph the function.

The graph crosses the -axis at 2 and 3, thus verifying the results found by factorization.

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=12 \c=36$

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 .

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

To verify, graph the function.

The graph crosses the -axis at -6, thus verifying the result found by factorization.

Compare your answer with the correct one above

Question

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

Answer

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,

$\b=10 \c=25$

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 .

$\x+5=0\rightarrow x=-5$

To verify, graph the function.

The graph crosses the -axis at -5, thus verifying the result found by factorization.

Compare your answer with the correct one above

Question

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

Answer

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

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

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

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Question

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

Answer

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

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

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

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Question

Find the zeros of $- 7 x^{2} + 14 x + 33$

Answer

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.

$x =\frac{ -14 \pm \sqrt{ 14 ^2 - 4 * -7 * 33 }}{ 2 * -7 }$

$x =\frac{ -14 \pm \sqrt{ 196 - -924 }}{ -14 }$

$x =\frac{ -14 \pm \sqrt{ 1120 }}{ -14 }$

$x =\frac{ -14 \pm 4 \sqrt{70} }{ -14 }$

Now we split this up into two equations.

$x =\frac{ -14 + 4 \sqrt{70} }{ -14 }$

$x =\frac{ -14 - 4 \sqrt{70} }{ -14 }$

So our zeros are at

$x = 1 + \frac{2 \sqrt{70}}{7}$

$x = - \frac{2 \sqrt{70}}{7} + 1$

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Question

Find the zeros of $9 x^{2} + 34 x - 99$

Answer

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.

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 9 * -99 }}{ 2 * 9 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - -3564 }}{ 18 }$

$x =\frac{ -34 \pm \sqrt{ 4720 }}{ 18 }$

$x =\frac{ -34 \pm 4 \sqrt{295} }{ 18 }$

Now we split this up into two equations.

$x =\frac{ -34 + 4 \sqrt{295} }{ 18 }$

$x =\frac{ -34 - 4 \sqrt{295} }{ 18 }$

So our zeros are at

$x = - \frac{17}{9} + \frac{2 \sqrt{295}}{9}$

$x = - \frac{2 \sqrt{295}}{9} - \frac{17}{9}$

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Question

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

Answer

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

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

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

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Question

Find the zeros of $- 5 x^{2} + 36 x + 76$

Answer

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.

$x =\frac{ -36 \pm \sqrt{ 36 ^2 - 4 * -5 * 76 }}{ 2 * -5 }$

$x =\frac{ -36 \pm \sqrt{ 1296 - -1520 }}{ -10 }$

$x =\frac{ -36 \pm \sqrt{ 2816 }}{ -10 }$

$x =\frac{ -36 \pm 16 \sqrt{11} }{ -10 }$

Now we split this up into two equations.

$x =\frac{ -36 + 16 \sqrt{11} }{ -10 }$

$x =\frac{ -36 - 16 \sqrt{11} }{ -10 }$

So our zeros are at

$x = \frac{18}{5} + \frac{8 \sqrt{11}}{5}$

$x = - \frac{8 \sqrt{11}}{5} + \frac{18}{5}$

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Question

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

Answer

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

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

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

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Question

Find the zeros of $x^{2} + 34 x + 24$

Answer

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.

$x =\frac{ -34 \pm \sqrt{ 34 ^2 - 4 * 1 * 24 }}{ 2 * 1 }$

$x =\frac{ -34 \pm \sqrt{ 1156 - 96 }}{ 2 }$

$x =\frac{ -34 \pm \sqrt{ 1060 }}{ 2 }$

$x =\frac{ -34 \pm 2 \sqrt{265} }{ 2 }$

Now we split this up into two equations.

$x =\frac{ -34 + 2 \sqrt{265} }{ 2 }$

$x =\frac{ -34 - 2 \sqrt{265} }{ 2 }$

So our zeros are at

$\x = -17 - \sqrt{265} \x = -17 + \sqrt{265}$

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