Algebra Examples

$(- 1, - 4)$ , $(0, 0)$ , $(2, - 10)$

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.

$- 4 = a {(- 1)}^{2} + b (- 1) + c, 0 = a {(0)}^{2} + b (0) + c, - 10 = a {(2)}^{2} + b (2) + c$

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 $c$ in $0 = a \cdot 0^{2} + c$ .

Tap for more steps...

Step 3.1.1

Rewrite the equation as $a \cdot 0^{2} + c = 0$ .

$a \cdot 0^{2} + c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.1.2

Simplify $a \cdot 0^{2} + c$ .

Tap for more steps...

Step 3.1.2.1

Simplify each term.

Tap for more steps...

Step 3.1.2.1.1

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

$a \cdot 0 + c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.1.2.1.2

Multiply $a$ by $0$ .

$0 + c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

$0 + c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.1.2.2

Add $0$ and $c$ .

$c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

$c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

$c = 0$

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

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2

Replace all occurrences of $c$ with $0$ in each equation.

Tap for more steps...

Step 3.2.1

Replace all occurrences of $c$ in $- 4 = a {(- 1)}^{2} + b (- 1) + c$ with $0$ .

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2

Simplify $- 4 = a {(- 1)}^{2} + b (- 1) + 0$ .

Tap for more steps...

Step 3.2.2.1

Simplify the left side.

Tap for more steps...

Step 3.2.2.1.1

Remove parentheses.

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2.2

Simplify the right side.

Tap for more steps...

Step 3.2.2.2.1

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

Tap for more steps...

Step 3.2.2.2.1.1

Add $a {(- 1)}^{2} + b (- 1)$ and $0$ .

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2.2.1.2

Simplify each term.

Tap for more steps...

Step 3.2.2.2.1.2.1

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

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2.2.1.2.2

Multiply $a$ by $1$ .

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

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2.2.1.2.3

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

$- 4 = a - 1 b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.2.2.1.2.4

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

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + c$

Step 3.2.3

Replace all occurrences of $c$ in $- 10 = a \cdot 2^{2} + b (2) + c$ with $0$ .

$- 10 = a \cdot 2^{2} + b (2) + 0$

$- 4 = a - b$

$c = 0$

Step 3.2.4

Simplify $- 10 = a \cdot 2^{2} + b (2) + 0$ .

Tap for more steps...

Step 3.2.4.1

Simplify the left side.

Tap for more steps...

Step 3.2.4.1.1

Remove parentheses.

$- 10 = a \cdot 2^{2} + b (2) + 0$

$- 4 = a - b$

$c = 0$

$- 10 = a \cdot 2^{2} + b (2) + 0$

$- 4 = a - b$

$c = 0$

Step 3.2.4.2

Simplify the right side.

Tap for more steps...

Step 3.2.4.2.1

Simplify $a \cdot 2^{2} + b (2) + 0$ .

Tap for more steps...

Step 3.2.4.2.1.1

Add $a \cdot 2^{2} + b (2)$ and $0$ .

$- 10 = a \cdot 2^{2} + b (2)$

$- 4 = a - b$

$c = 0$

Step 3.2.4.2.1.2

Simplify each term.

Tap for more steps...

Step 3.2.4.2.1.2.1

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

$- 10 = a \cdot 4 + b (2)$

$- 4 = a - b$

$c = 0$

Step 3.2.4.2.1.2.2

Move $4$ to the left of $a$ .

$- 10 = 4 \cdot a + b (2)$

$- 4 = a - b$

$c = 0$

Step 3.2.4.2.1.2.3

Move $2$ to the left of $b$ .

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

$- 10 = 4 a + 2 b$

$- 4 = a - b$

$c = 0$

Step 3.3

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

Tap for more steps...

Step 3.3.1

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

$a - b = - 4$

$- 10 = 4 a + 2 b$

$c = 0$

Step 3.3.2

Add $b$ to both sides of the equation.

$a = - 4 + b$

$- 10 = 4 a + 2 b$

$c = 0$

$a = - 4 + b$

$- 10 = 4 a + 2 b$

$c = 0$

Step 3.4

Replace all occurrences of $a$ with $- 4 + b$ in each equation.

Tap for more steps...

Step 3.4.1

Replace all occurrences of $a$ in $- 10 = 4 a + 2 b$ with $- 4 + b$ .

$- 10 = 4 (- 4 + b) + 2 b$

$a = - 4 + b$

$c = 0$

Step 3.4.2

Simplify the right side.

Tap for more steps...

Step 3.4.2.1

Simplify $4 (- 4 + b) + 2 b$ .

Tap for more steps...

Step 3.4.2.1.1

Simplify each term.

Tap for more steps...

Step 3.4.2.1.1.1

Apply the distributive property.

$- 10 = 4 \cdot - 4 + 4 b + 2 b$

$a = - 4 + b$

$c = 0$

Step 3.4.2.1.1.2

Multiply $4$ by $- 4$ .

$- 10 = - 16 + 4 b + 2 b$

$a = - 4 + b$

$c = 0$

$- 10 = - 16 + 4 b + 2 b$

$a = - 4 + b$

$c = 0$

Step 3.4.2.1.2

Add $4 b$ and $2 b$ .

$- 10 = - 16 + 6 b$

$a = - 4 + b$

$c = 0$

$- 10 = - 16 + 6 b$

$a = - 4 + b$

$c = 0$

$- 10 = - 16 + 6 b$

$a = - 4 + b$

$c = 0$

$- 10 = - 16 + 6 b$

$a = - 4 + b$

$c = 0$

Step 3.5

Solve for $b$ in $- 10 = - 16 + 6 b$ .

Tap for more steps...

Step 3.5.1

Rewrite the equation as $- 16 + 6 b = - 10$ .

$- 16 + 6 b = - 10$

$a = - 4 + b$

$c = 0$

Step 3.5.2

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

Tap for more steps...

Step 3.5.2.1

Add $16$ to both sides of the equation.

$6 b = - 10 + 16$

$a = - 4 + b$

$c = 0$

Step 3.5.2.2

Add $- 10$ and $16$ .

$6 b = 6$

$a = - 4 + b$

$c = 0$

$6 b = 6$

$a = - 4 + b$

$c = 0$

Step 3.5.3

Divide each term in $6 b = 6$ by $6$ and simplify.

Tap for more steps...

Step 3.5.3.1

Divide each term in $6 b = 6$ by $6$ .

$\frac{6 b}{6} = \frac{6}{6}$

$a = - 4 + b$

$c = 0$

Step 3.5.3.2

Simplify the left side.

Tap for more steps...

Step 3.5.3.2.1

Cancel the common factor of $6$ .

Tap for more steps...

Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{6 b}{6} = \frac{6}{6}$

$a = - 4 + b$

$c = 0$

Step 3.5.3.2.1.2

Divide $b$ by $1$ .

$b = \frac{6}{6}$

$a = - 4 + b$

$c = 0$

$b = \frac{6}{6}$

$a = - 4 + b$

$c = 0$

$b = \frac{6}{6}$

$a = - 4 + b$

$c = 0$

Step 3.5.3.3

Simplify the right side.

Tap for more steps...

Step 3.5.3.3.1

Divide $6$ by $6$ .

$b = 1$

$a = - 4 + b$

$c = 0$

$b = 1$

$a = - 4 + b$

$c = 0$

$b = 1$

$a = - 4 + b$

$c = 0$

$b = 1$

$a = - 4 + b$

$c = 0$

Step 3.6

Replace all occurrences of $b$ with $1$ in each equation.

Tap for more steps...

Step 3.6.1

Replace all occurrences of $b$ in $a = - 4 + b$ with $1$ .

$a = - 4 + 1$

$b = 1$

$c = 0$

Step 3.6.2

Simplify $a = - 4 + 1$ .

Tap for more steps...

Step 3.6.2.1

Simplify the left side.

Tap for more steps...

Step 3.6.2.1.1

Remove parentheses.

$a = - 4 + 1$

$b = 1$

$c = 0$

$a = - 4 + 1$

$b = 1$

$c = 0$

Step 3.6.2.2

Simplify the right side.

Tap for more steps...

Step 3.6.2.2.1

Add $- 4$ and $1$ .

$a = - 3$

$b = 1$

$c = 0$

$a = - 3$

$b = 1$

$c = 0$

$a = - 3$

$b = 1$

$c = 0$

$a = - 3$

$b = 1$

$c = 0$

Step 3.7

List all of the solutions.

$a = - 3, b = 1, c = 0$

Step 4

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

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

Step 5