Find the x and y Intercepts f(x)=-3x^4+x^3-2x^2+x-1

$f (x) = - 3 x^{4} + x^{3} - 2 x^{2} + x - 1$

Find the x-intercepts.

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

$0 = - 3 x^{4} + x^{3} - 2 x^{2} + x - 1$

Solve the equation.

Remove parentheses.

$0 = - 3 x^{4} + x^{3} - 2 x^{2} + x - 1$

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

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

x-intercept(s): $None$

Find the y-intercepts.

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

$y = - 3 {(0)}^{4} + {(0)}^{3} - 2 {(0)}^{2} + 0 - 1$

Solve the equation.

Remove parentheses.

$y = - 3 {(0)}^{4} + {(0)}^{3} - 2 {(0)}^{2} + 0 - 1$

Remove parentheses.

$y = - 3 \cdot 0^{4} + {(0)}^{3} - 2 {(0)}^{2} + 0 - 1$

Remove parentheses.

$y = - 3 \cdot 0^{4} + 0^{3} - 2 {(0)}^{2} + 0 - 1$

Remove parentheses.

$y = - 3 \cdot 0^{4} + 0^{3} - 2 \cdot 0^{2} + 0 - 1$

Remove parentheses.

$y = - 3 {(0)}^{4} + {(0)}^{3} - 2 {(0)}^{2} + 0 - 1$

Simplify $- 3 {(0)}^{4} + {(0)}^{3} - 2 {(0)}^{2} + 0 - 1$ .

Simplify each term.

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

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

Multiply $- 3$ by $0$ .

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

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

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

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

$y = 0 + 0 - 2 \cdot 0 + 0 - 1$

Multiply $- 2$ by $0$ .

$y = 0 + 0 + 0 + 0 - 1$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 + 0 - 1$

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): $(0, - 1)$

List the intersections.

x-intercept(s): $None$

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

Find the x and y Intercepts f(x)=-3x^4+x^3-2x^2+x-1

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