Find the x and y Intercepts f(x)=-x^3+x^2+x-1

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

Step 1

Find the x-intercepts.

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

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

Solve the equation.

Rewrite the equation as $- x^{3} + x^{2} + x - 1 = 0$ .

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

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

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

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

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

Rewrite $1$ as $1^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

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

Remove unnecessary parentheses.

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

Combine exponents.

Factor $- 1$ out of $- x$ .

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

Rewrite $1$ as $- 1 (- 1)$ .

$(- (x) - 1 \cdot - 1) (x + 1) (x - 1) = 0$

Factor $- 1$ out of $- (x) - 1 (- 1)$ .

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

Rewrite $- (x - 1)$ as $- 1 (x - 1)$ .

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

Raise $x - 1$ to the power of $1$ .

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

Raise $x - 1$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

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

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

$x + 1 = 0$

Set ${(x - 1)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(x - 1)}^{2}$ equal to $0$ .

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

Solve ${(x - 1)}^{2} = 0$ for $x$ .

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = 1, - 1$

x-intercept(s) in point form.

x-intercept(s): $(1, 0), (- 1, 0)$

Step 2

Find the y-intercepts.

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

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

Solve the equation.

Remove parentheses.

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

Remove parentheses.

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

Remove parentheses.

$y = - 0^{3} + 0^{2} + 0 - 1$

Remove parentheses.

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

Simplify $- {(0)}^{3} + {(0)}^{2} + 0 - 1$ .

Simplify each term.

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

$y = - 0 + {(0)}^{2} + 0 - 1$

Multiply $- 1$ by $0$ .

$y = 0 + {(0)}^{2} + 0 - 1$

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

$y = 0 + 0 + 0 - 1$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 - 1$

Add $0$ and $0$ .

$y = 0 - 1$

Subtract $1$ from $0$ .

$y = - 1$

y-intercept(s) in point form.

y-intercept(s): $(0, - 1)$

Step 3

List the intersections.

x-intercept(s): $(1, 0), (- 1, 0)$

y-intercept(s): $(0, - 1)$

Step 4

Find the x and y Intercepts f(x)=-x^3+x^2+x-1

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