Graph y=-(2x+3)(x-1)^2

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

Find the point at $x = - 1$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = - 2 \cdot - 1 {((- 1) - 1)}^{2} - 3 {((- 1) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $- 2$ by $- 1$ .

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

Subtract $1$ from $- 1$ .

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

Raise $- 2$ to the power of $2$ .

$f (- 1) = 2 \cdot 4 - 3 {((- 1) - 1)}^{2}$

Multiply $2$ by $4$ .

$f (- 1) = 8 - 3 {((- 1) - 1)}^{2}$

Subtract $1$ from $- 1$ .

$f (- 1) = 8 - 3 {(- 2)}^{2}$

Raise $- 2$ to the power of $2$ .

$f (- 1) = 8 - 3 \cdot 4$

Multiply $- 3$ by $4$ .

$f (- 1) = 8 - 12$

Subtract $12$ from $8$ .

$f (- 1) = - 4$

The final answer is $- 4$ .

$- 4$

Convert $- 4$ to decimal.

$y = - 4$

Find the point at $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

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

Simplify the result.

Simplify each term.

Multiply $- 2$ by $0$ .

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

Subtract $1$ from $0$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply $0$ by $1$ .

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

Subtract $1$ from $0$ .

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

Raise $- 1$ to the power of $2$ .

$f (0) = 0 - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$f (0) = 0 - 3$

Subtract $3$ from $0$ .

$f (0) = - 3$

The final answer is $- 3$ .

$- 3$

Convert $- 3$ to decimal.

$y = - 3$

Find the point at $x = 1$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = - 2 \cdot 1 {((1) - 1)}^{2} - 3 {((1) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $- 2$ by $1$ .

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

Subtract $1$ from $1$ .

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

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

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

Multiply $- 2$ by $0$ .

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

Subtract $1$ from $1$ .

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

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

$f (1) = 0 - 3 \cdot 0$

Multiply $- 3$ by $0$ .

$f (1) = 0 + 0$

Add $0$ and $0$ .

$f (1) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 2$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = - 2 \cdot 2 {((2) - 1)}^{2} - 3 {((2) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $- 2$ by $2$ .

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

Subtract $1$ from $2$ .

$f (2) = - 4 \cdot 1^{2} - 3 {((2) - 1)}^{2}$

One to any power is one.

$f (2) = - 4 \cdot 1 - 3 {((2) - 1)}^{2}$

Multiply $- 4$ by $1$ .

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

Subtract $1$ from $2$ .

$f (2) = - 4 - 3 \cdot 1^{2}$

One to any power is one.

$f (2) = - 4 - 3 \cdot 1$

Multiply $- 3$ by $1$ .

$f (2) = - 4 - 3$

Subtract $3$ from $- 4$ .

$f (2) = - 7$

The final answer is $- 7$ .

$- 7$

Convert $- 7$ to decimal.

$y = - 7$

Find the point at $x = 3$ .

Replace the variable $x$ with $3$ in the expression.

$f (3) = - 2 \cdot 3 {((3) - 1)}^{2} - 3 {((3) - 1)}^{2}$

Simplify the result.

Simplify each term.

Multiply $- 2$ by $3$ .

$f (3) = - 6 {((3) - 1)}^{2} - 3 {((3) - 1)}^{2}$

Subtract $1$ from $3$ .

$f (3) = - 6 \cdot 2^{2} - 3 {((3) - 1)}^{2}$

Raise $2$ to the power of $2$ .

$f (3) = - 6 \cdot 4 - 3 {((3) - 1)}^{2}$

Multiply $- 6$ by $4$ .

$f (3) = - 24 - 3 {((3) - 1)}^{2}$

Subtract $1$ from $3$ .

$f (3) = - 24 - 3 \cdot 2^{2}$

Raise $2$ to the power of $2$ .

$f (3) = - 24 - 3 \cdot 4$

Multiply $- 3$ by $4$ .

$f (3) = - 24 - 12$

Subtract $12$ from $- 24$ .

$f (3) = - 36$

The final answer is $- 36$ .

$- 36$

Convert $- 36$ to decimal.

$y = - 36$

The cubic function can be graphed using the function behavior and the points.

$\begin{array}{c} x & y \\ - 1 & - 4 \\ 0 & - 3 \\ 1 & 0 \\ 2 & - 7 \\ 3 & - 36 \end{array}$

The cubic function can be graphed using the function behavior and the selected points.

Rises to the left and falls to the right

$\begin{array}{c} x & y \\ - 1 & - 4 \\ 0 & - 3 \\ 1 & 0 \\ 2 & - 7 \\ 3 & - 36 \end{array}$

Graph y=-(2x+3)(x-1)^2

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