Precalculus Examples

$f (x) = 4 x^{4} - 17 x^{2} + 4$

Step 1

Find the x-intercepts.

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Step 1.1

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 4 x^{4} - 17 x^{2} + 4$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $4 x^{4} - 17 x^{2} + 4 = 0$ .

$4 x^{4} - 17 x^{2} + 4 = 0$

Step 1.2.2

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$4 u^{2} - 17 u + 4 = 0$

$u = x^{2}$

Step 1.2.3

Factor by grouping.

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Step 1.2.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 4 = 16$ and whose sum is $b = - 17$ .

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Step 1.2.3.1.1

Factor $- 17$ out of $- 17 u$ .

$4 u^{2} - 17 u + 4 = 0$

Step 1.2.3.1.2

Rewrite $- 17$ as $- 1$ plus $- 16$

$4 u^{2} + (- 1 - 16) u + 4 = 0$

Step 1.2.3.1.3

Apply the distributive property.

$4 u^{2} - 1 u - 16 u + 4 = 0$

Step 1.2.3.2

Factor out the greatest common factor from each group.

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Step 1.2.3.2.1

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) - 16 u + 4 = 0$

Step 1.2.3.2.2

Factor out the greatest common factor (GCF) from each group.

$u (4 u - 1) - 4 (4 u - 1) = 0$

Step 1.2.3.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u - 4) = 0$

Step 1.2.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 u - 1 = 0$

$u - 4 = 0$

Step 1.2.5

Set $4 u - 1$ equal to $0$ and solve for $u$ .

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Step 1.2.5.1

Set $4 u - 1$ equal to $0$ .

$4 u - 1 = 0$

Step 1.2.5.2

Solve $4 u - 1 = 0$ for $u$ .

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Step 1.2.5.2.1

Add $1$ to both sides of the equation.

$4 u = 1$

Step 1.2.5.2.2

Divide each term in $4 u = 1$ by $4$ and simplify.

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Step 1.2.5.2.2.1

Divide each term in $4 u = 1$ by $4$ .

$\frac{4 u}{4} = \frac{1}{4}$

Step 1.2.5.2.2.2

Simplify the left side.

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Step 1.2.5.2.2.2.1

Cancel the common factor of $4$ .

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Step 1.2.5.2.2.2.1.1

Cancel the common factor.

$\frac{4 u}{4} = \frac{1}{4}$

Step 1.2.5.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{4}$

Step 1.2.6

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 1.2.6.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 1.2.6.2

Add $4$ to both sides of the equation.

$u = 4$

Step 1.2.7

The final solution is all the values that make $(4 u - 1) (u - 4) = 0$ true.

$u = \frac{1}{4}, 4$

Step 1.2.8

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = \frac{1}{4}$

${(x^{2})}^{1} = 4$

Step 1.2.9

Solve the first equation for $x$ .

$x^{2} = \frac{1}{4}$

Step 1.2.10

Solve the equation for $x$ .

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Step 1.2.10.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{\frac{1}{4}}$

Step 1.2.10.2

Simplify $\pm \sqrt{\frac{1}{4}}$ .

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Step 1.2.10.2.1

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$x = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Step 1.2.10.2.2

Any root of $1$ is $1$ .

$x = \pm \frac{1}{\sqrt{4}}$

Step 1.2.10.2.3

Simplify the denominator.

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Step 1.2.10.2.3.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \frac{1}{\sqrt{2^{2}}}$

Step 1.2.10.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm \frac{1}{2}$

Step 1.2.10.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.10.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x = \frac{1}{2}$

Step 1.2.10.3.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \frac{1}{2}$

Step 1.2.10.3.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{1}{2}, - \frac{1}{2}$

Step 1.2.11

Solve the second equation for $x$ .

${(x^{2})}^{1} = 4$

Step 1.2.12

Solve the equation for $x$ .

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Step 1.2.12.1

Remove parentheses.

$x^{2} = 4$

Step 1.2.12.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{4}$

Step 1.2.12.3

Simplify $\pm \sqrt{4}$ .

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Step 1.2.12.3.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 1.2.12.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

Step 1.2.12.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.2.12.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 2$

Step 1.2.12.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 2$

Step 1.2.12.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 2, - 2$

Step 1.2.13

The solution to $4 x^{4} - 17 x^{2} + 4 = 0$ is $x = \frac{1}{2}, - \frac{1}{2}, 2, - 2$ .

$x = \frac{1}{2}, - \frac{1}{2}, 2, - 2$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(\frac{1}{2}, 0), (- \frac{1}{2}, 0), (2, 0), (- 2, 0)$

Step 2

Find the y-intercepts.

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Step 2.1

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 4 {(0)}^{4} - 17 {(0)}^{2} + 4$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = 4 \cdot 0^{4} - 17 {(0)}^{2} + 4$

Step 2.2.2

Remove parentheses.

$y = 4 \cdot 0^{4} - 17 \cdot 0^{2} + 4$

Step 2.2.3

Remove parentheses.

$y = 4 {(0)}^{4} - 17 {(0)}^{2} + 4$

Step 2.2.4

Simplify $4 {(0)}^{4} - 17 {(0)}^{2} + 4$ .

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Step 2.2.4.1

Simplify each term.

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Step 2.2.4.1.1

Raising $0$ to any positive power yields $0$ .

$y = 4 \cdot 0 - 17 {(0)}^{2} + 4$

Step 2.2.4.1.2

Multiply $4$ by $0$ .

$y = 0 - 17 {(0)}^{2} + 4$

Step 2.2.4.1.3

Raising $0$ to any positive power yields $0$ .

$y = 0 - 17 \cdot 0 + 4$

Step 2.2.4.1.4

Multiply $- 17$ by $0$ .

$y = 0 + 0 + 4$

Step 2.2.4.2

Simplify by adding numbers.

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Step 2.2.4.2.1

Add $0$ and $0$ .

$y = 0 + 4$

Step 2.2.4.2.2

Add $0$ and $4$ .

$y = 4$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, 4)$

Step 3

List the intersections.

x-intercept(s): $(\frac{1}{2}, 0), (- \frac{1}{2}, 0), (2, 0), (- 2, 0)$

y-intercept(s): $(0, 4)$

Step 4