Algebra Examples

$(- 3, 0)$ , $(- 1, 4)$ , $(1, 0)$

Step 1

Use the standard form of a quadratic equation $y = a x^{2} + b x + c$ as the starting point for finding the equation through the three points.

$y = a x^{2} + b x + c$

Step 2

Create a system of equations by substituting the $x$ and $y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

$0 = a {(- 3)}^{2} + b (- 3) + c, 4 = a {(- 1)}^{2} + b (- 1) + c, 0 = a {(1)}^{2} + b (1) + c$

Step 3

Solve the system of equations to find the values of $a$ , $b$ , and $c$ .

Tap for more steps...

Step 3.1

Solve for $a$ in $0 = a + b + c$ .

Tap for more steps...

Step 3.1.1

Rewrite the equation as $a + b + c = 0$ .

$a + b + c = 0$

$0 = a {(- 3)}^{2} + b (- 3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.1.2

Move all terms not containing $a$ to the right side of the equation.

Tap for more steps...

Step 3.1.2.1

Subtract $b$ from both sides of the equation.

$a + c = - b$

$0 = a {(- 3)}^{2} + b (- 3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.1.2.2

Subtract $c$ from both sides of the equation.

$a = - b - c$

$0 = a {(- 3)}^{2} + b (- 3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$a = - b - c$

$0 = a {(- 3)}^{2} + b (- 3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$a = - b - c$

$0 = a {(- 3)}^{2} + b (- 3) + c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2

Replace all occurrences of $a$ with $- b - c$ in each equation.

Tap for more steps...

Step 3.2.1

Replace all occurrences of $a$ in $0 = a {(- 3)}^{2} + b (- 3) + c$ with $- b - c$ .

$0 = (- b - c) {(- 3)}^{2} + b (- 3) + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.1

Simplify $(- b - c) {(- 3)}^{2} + b (- 3) + c$ .

Tap for more steps...

Step 3.2.2.1.1

Simplify each term.

Tap for more steps...

Step 3.2.2.1.1.1

Raise $- 3$ to the power of $2$ .

$0 = (- b - c) \cdot 9 + b (- 3) + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.1.2

Apply the distributive property.

$0 = - b \cdot 9 - c \cdot 9 + b (- 3) + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.1.3

Multiply $9$ by $- 1$ .

$0 = - 9 b - c \cdot 9 + b (- 3) + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.1.4

Multiply $9$ by $- 1$ .

$0 = - 9 b - 9 c + b (- 3) + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.1.5

Move $- 3$ to the left of $b$ .

$0 = - 9 b - 9 c - 3 b + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$0 = - 9 b - 9 c - 3 b + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.2

Simplify by adding terms.

Tap for more steps...

Step 3.2.2.1.2.1

Subtract $3 b$ from $- 9 b$ .

$0 = - 12 b - 9 c + c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.2.1.2.2

Add $- 9 c$ and $c$ .

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = a {(- 1)}^{2} + b (- 1) + c$

Step 3.2.3

Replace all occurrences of $a$ in $4 = a {(- 1)}^{2} + b (- 1) + c$ with $- b - c$ .

$4 = (- b - c) {(- 1)}^{2} + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4

Simplify the right side.

Tap for more steps...

Step 3.2.4.1

Simplify $(- b - c) {(- 1)}^{2} + b (- 1) + c$ .

Tap for more steps...

Step 3.2.4.1.1

Simplify each term.

Tap for more steps...

Step 3.2.4.1.1.1

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

$4 = (- b - c) \cdot 1 + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.1.2

Multiply $- b - c$ by $1$ .

$4 = - b - c + b (- 1) + c$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.1.3

Move $- 1$ to the left of $b$ .

$4 = - b - c - 1 b + c$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.1.4

Rewrite $- 1 b$ as $- b$ .

$4 = - b - c - b + c$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - b - c - b + c$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.2

Simplify by adding terms.

Tap for more steps...

Step 3.2.4.1.2.1

Combine the opposite terms in $- b - c - b + c$ .

Tap for more steps...

Step 3.2.4.1.2.1.1

Add $- c$ and $c$ .

$4 = - b - b + 0$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.2.1.2

Add $- b - b$ and $0$ .

$4 = - b - b$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - b - b$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.2.4.1.2.2

Subtract $b$ from $- b$ .

$4 = - 2 b$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - 2 b$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - 2 b$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - 2 b$

$0 = - 12 b - 8 c$

$a = - b - c$

$4 = - 2 b$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.3

Solve for $b$ in $4 = - 2 b$ .

Tap for more steps...

Step 3.3.1

Rewrite the equation as $- 2 b = 4$ .

$- 2 b = 4$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.3.2

Divide each term in $- 2 b = 4$ by $- 2$ and simplify.

Tap for more steps...

Step 3.3.2.1

Divide each term in $- 2 b = 4$ by $- 2$ .

$\frac{- 2 b}{- 2} = \frac{4}{- 2}$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.3.2.2

Simplify the left side.

Tap for more steps...

Step 3.3.2.2.1

Cancel the common factor of $- 2$ .

Tap for more steps...

Step 3.3.2.2.1.1

Cancel the common factor.

$\frac{- 2 b}{- 2} = \frac{4}{- 2}$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.3.2.2.1.2

Divide $b$ by $1$ .

$b = \frac{4}{- 2}$

$0 = - 12 b - 8 c$

$a = - b - c$

$b = \frac{4}{- 2}$

$0 = - 12 b - 8 c$

$a = - b - c$

$b = \frac{4}{- 2}$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.3.2.3

Simplify the right side.

Tap for more steps...

Step 3.3.2.3.1

Divide $4$ by $- 2$ .

$b = - 2$

$0 = - 12 b - 8 c$

$a = - b - c$

$b = - 2$

$0 = - 12 b - 8 c$

$a = - b - c$

$b = - 2$

$0 = - 12 b - 8 c$

$a = - b - c$

$b = - 2$

$0 = - 12 b - 8 c$

$a = - b - c$

Step 3.4

Replace all occurrences of $b$ with $- 2$ in each equation.

Tap for more steps...

Step 3.4.1

Replace all occurrences of $b$ in $0 = - 12 b - 8 c$ with $- 2$ .

$0 = - 12 \cdot - 2 - 8 c$

$b = - 2$

$a = - b - c$

Step 3.4.2

Simplify the right side.

Tap for more steps...

Step 3.4.2.1

Multiply $- 12$ by $- 2$ .

$0 = 24 - 8 c$

$b = - 2$

$a = - b - c$

$0 = 24 - 8 c$

$b = - 2$

$a = - b - c$

Step 3.4.3

Replace all occurrences of $b$ in $a = - b - c$ with $- 2$ .

$a = - (- 2) - c$

$0 = 24 - 8 c$

$b = - 2$

Step 3.4.4

Simplify the right side.

Tap for more steps...

Step 3.4.4.1

Multiply $- 1$ by $- 2$ .

$a = 2 - c$

$0 = 24 - 8 c$

$b = - 2$

$a = 2 - c$

$0 = 24 - 8 c$

$b = - 2$

$a = 2 - c$

$0 = 24 - 8 c$

$b = - 2$

Step 3.5

Solve for $c$ in $0 = 24 - 8 c$ .

Tap for more steps...

Step 3.5.1

Rewrite the equation as $24 - 8 c = 0$ .

$24 - 8 c = 0$

$a = 2 - c$

$b = - 2$

Step 3.5.2

Subtract $24$ from both sides of the equation.

$- 8 c = - 24$

$a = 2 - c$

$b = - 2$

Step 3.5.3

Divide each term in $- 8 c = - 24$ by $- 8$ and simplify.

Tap for more steps...

Step 3.5.3.1

Divide each term in $- 8 c = - 24$ by $- 8$ .

$\frac{- 8 c}{- 8} = \frac{- 24}{- 8}$

$a = 2 - c$

$b = - 2$

Step 3.5.3.2

Simplify the left side.

Tap for more steps...

Step 3.5.3.2.1

Cancel the common factor of $- 8$ .

Tap for more steps...

Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{- 8 c}{- 8} = \frac{- 24}{- 8}$

$a = 2 - c$

$b = - 2$

Step 3.5.3.2.1.2

Divide $c$ by $1$ .

$c = \frac{- 24}{- 8}$

$a = 2 - c$

$b = - 2$

$c = \frac{- 24}{- 8}$

$a = 2 - c$

$b = - 2$

$c = \frac{- 24}{- 8}$

$a = 2 - c$

$b = - 2$

Step 3.5.3.3

Simplify the right side.

Tap for more steps...

Step 3.5.3.3.1

Divide $- 24$ by $- 8$ .

$c = 3$

$a = 2 - c$

$b = - 2$

$c = 3$

$a = 2 - c$

$b = - 2$

$c = 3$

$a = 2 - c$

$b = - 2$

$c = 3$

$a = 2 - c$

$b = - 2$

Step 3.6

Replace all occurrences of $c$ with $3$ in each equation.

Tap for more steps...

Step 3.6.1

Replace all occurrences of $c$ in $a = 2 - c$ with $3$ .

$a = 2 - (3)$

$c = 3$

$b = - 2$

Step 3.6.2

Simplify the right side.

Tap for more steps...

Step 3.6.2.1

Simplify $2 - (3)$ .

Tap for more steps...

Step 3.6.2.1.1

Multiply $- 1$ by $3$ .

$a = 2 - 3$

$c = 3$

$b = - 2$

Step 3.6.2.1.2

Subtract $3$ from $2$ .

$a = - 1$

$c = 3$

$b = - 2$

$a = - 1$

$c = 3$

$b = - 2$

$a = - 1$

$c = 3$

$b = - 2$

$a = - 1$

$c = 3$

$b = - 2$

Step 3.7

List all of the solutions.

$a = - 1, c = 3, b = - 2$

Step 4

Substitute the actual values of $a$ , $b$ , and $c$ into the formula for a quadratic equation to find the resulting equation.

$y = - x^{2} - 2 x + 3$

Step 5