Precalculus Examples

$f (x) = 2 x^{2} + 5 x - 3$

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 $x$ .

$0 = 2 x^{2} + 5 x - 3$

Step 1.2

Solve the equation.

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Step 1.2.1

Rewrite the equation as $2 x^{2} + 5 x - 3 = 0$ .

$2 x^{2} + 5 x - 3 = 0$

Step 1.2.2

Factor by grouping.

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Step 1.2.2.1

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 $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

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Step 1.2.2.1.1

Factor $5$ out of $5 x$ .

$2 x^{2} + 5 (x) - 3 = 0$

Step 1.2.2.1.2

Rewrite $5$ as $- 1$ plus $6$

$2 x^{2} + (- 1 + 6) x - 3 = 0$

Step 1.2.2.1.3

Apply the distributive property.

$2 x^{2} - 1 x + 6 x - 3 = 0$

Step 1.2.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.2.1

Group the first two terms and the last two terms.

$(2 x^{2} - 1 x) + 6 x - 3 = 0$

Step 1.2.2.2.2

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

$x (2 x - 1) + 3 (2 x - 1) = 0$

Step 1.2.2.3

Factor the polynomial by factoring out the greatest common factor, $2 x - 1$ .

$(2 x - 1) (x + 3) = 0$

Step 1.2.3

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

$2 x - 1 = 0$

$x + 3 = 0$

Step 1.2.4

Set $2 x - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Step 1.2.4.2

Solve $2 x - 1 = 0$ for $x$ .

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Step 1.2.4.2.1

Add $1$ to both sides of the equation.

$2 x = 1$

Step 1.2.4.2.2

Divide each term in $2 x = 1$ by $2$ and simplify.

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Step 1.2.4.2.2.1

Divide each term in $2 x = 1$ by $2$ .

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

Step 1.2.4.2.2.2

Simplify the left side.

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Step 1.2.4.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.4.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.4.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 1.2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 1.2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 1.2.6

The final solution is all the values that make $(2 x - 1) (x + 3) = 0$ true.

$x = \frac{1}{2}, - 3$

Step 1.3

x-intercept(s) in point form.

x-intercept(s): $(\frac{1}{2}, 0), (- 3, 0)$

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 $y$ .

$y = 2 {(0)}^{2} + 5 (0) - 3$

Step 2.2

Solve the equation.

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Step 2.2.1

Remove parentheses.

$y = 2 \cdot 0^{2} + 5 (0) - 3$

Step 2.2.2

Remove parentheses.

$y = 2 {(0)}^{2} + 5 (0) - 3$

Step 2.2.3

Simplify $2 {(0)}^{2} + 5 (0) - 3$ .

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Step 2.2.3.1

Simplify each term.

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Step 2.2.3.1.1

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

$y = 2 \cdot 0 + 5 (0) - 3$

Step 2.2.3.1.2

Multiply $2$ by $0$ .

$y = 0 + 5 (0) - 3$

Step 2.2.3.1.3

Multiply $5$ by $0$ .

$y = 0 + 0 - 3$

Step 2.2.3.2

Simplify by adding and subtracting.

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Step 2.2.3.2.1

Add $0$ and $0$ .

$y = 0 - 3$

Step 2.2.3.2.2

Subtract $3$ from $0$ .

$y = - 3$

Step 2.3

y-intercept(s) in point form.

y-intercept(s): $(0, - 3)$

Step 3

List the intersections.

x-intercept(s): $(\frac{1}{2}, 0), (- 3, 0)$

y-intercept(s): $(0, - 3)$

Step 4