Precalculus Examples

(0, 9)

$(0, 9)$ ,

(5, 4)

$(5, 4)$ ,

(1, 4)

$(1, 4)$

Step 1

Use the standard form of a quadratic equation

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

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

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

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

Step 2

Create a system of equations by substituting the

x

$x$ and

y

$y$ values of each point into the standard formula of a quadratic equation to create the three equation system.

9 = a {(0)}^{2} + b (0) + c, 4 = a {(5)}^{2} + b (5) + c, 4 = a {(1)}^{2} + b (1) + c

$9 = a {(0)}^{2} + b (0) + c, 4 = a {(5)}^{2} + b (5) + c, 4 = a {(1)}^{2} + b (1) + c$

Step 3

Solve the system of equations to find the values of

a

$a$ ,

b

$b$ , and

c

$c$ .

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Step 3.1

Solve for

c

$c$ in

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

$9 = a \cdot 0^{2} + c$ .

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Step 3.1.1

Rewrite the equation as

a \cdot 0^{2} + c = 9

$a \cdot 0^{2} + c = 9$ .

a \cdot 0^{2} + c = 9

$a \cdot 0^{2} + c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

Step 3.1.2

Simplify

a \cdot 0^{2} + c

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

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Step 3.1.2.1

Simplify each term.

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Step 3.1.2.1.1

Raising

0

$0$ to any positive power yields

0

$0$ .

a \cdot 0 + c = 9

$a \cdot 0 + c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

Step 3.1.2.1.2

Multiply

a

$a$ by

0

$0$ .

0 + c = 9

$0 + c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

0 + c = 9

$0 + c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

Step 3.1.2.2

Add

0

$0$ and

c

$c$ .

c = 9

$c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

c = 9

$c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

c = 9

$c = 9$

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$

4 = a + b + c

$4 = a + b + c$

Step 3.2

Replace all occurrences of

c

$c$ with

9

$9$ in each equation.

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Step 3.2.1

Replace all occurrences of

c

$c$ in

4 = a \cdot 5^{2} + b (5) + c

$4 = a \cdot 5^{2} + b (5) + c$ with

9

$9$ .

4 = a \cdot 5^{2} + b (5) + 9

$4 = a \cdot 5^{2} + b (5) + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

Step 3.2.2

Simplify

4 = a \cdot 5^{2} + b (5) + 9

$4 = a \cdot 5^{2} + b (5) + 9$ .

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Step 3.2.2.1

Simplify the left side.

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Step 3.2.2.1.1

Remove parentheses.

4 = a \cdot 5^{2} + b (5) + 9

$4 = a \cdot 5^{2} + b (5) + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

4 = a \cdot 5^{2} + b (5) + 9

$4 = a \cdot 5^{2} + b (5) + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

Step 3.2.2.2

Simplify the right side.

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Step 3.2.2.2.1

Simplify each term.

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Step 3.2.2.2.1.1

Raise

5

$5$ to the power of

2

$2$ .

4 = a \cdot 25 + b (5) + 9

$4 = a \cdot 25 + b (5) + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

Step 3.2.2.2.1.2

Move

25

$25$ to the left of

a

$a$ .

4 = 25 \cdot a + b (5) + 9

$4 = 25 \cdot a + b (5) + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

Step 3.2.2.2.1.3

Move

5

$5$ to the left of

b

$b$ .

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + c

$4 = a + b + c$

Step 3.2.3

Replace all occurrences of

c

$c$ in

4 = a + b + c

$4 = a + b + c$ with

9

$9$ .

4 = a + b + 9

$4 = a + b + 9$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

Step 3.2.4

Simplify the left side.

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Step 3.2.4.1

Remove parentheses.

4 = a + b + 9

$4 = a + b + 9$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + 9

$4 = a + b + 9$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

c = 9

$c = 9$

4 = a + b + 9

$4 = a + b + 9$

4 = 25 a + 5 b + 9

$4 = 25 a + 5 b + 9$

$c = 9$

Step 3.3

Solve for

$a$ in

$4 = a + b + 9$ .

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Step 3.3.1

Rewrite the equation as

$a + b + 9 = 4$ .

$a + b + 9 = 4$

$4 = 25 a + 5 b + 9$

$c = 9$

Step 3.3.2

Move all terms not containing

$a$ to the right side of the equation.

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Step 3.3.2.1

Subtract

$b$ from both sides of the equation.

$a + 9 = 4 - b$

$4 = 25 a + 5 b + 9$

$c = 9$

Step 3.3.2.2

Subtract

$9$ from both sides of the equation.

$a = 4 - b - 9$

$4 = 25 a + 5 b + 9$

$c = 9$

Step 3.3.2.3

Subtract

$9$ from

$4$ .

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

$a = - b - 5$

$4 = 25 a + 5 b + 9$

$c = 9$

Step 3.4

Replace all occurrences of

$a$ with

$- b - 5$ in each equation.

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Step 3.4.1

Replace all occurrences of

$a$ in

$4 = 25 a + 5 b + 9$ with

$- b - 5$ .

$4 = 25 (- b - 5) + 5 b + 9$

$a = - b - 5$

$c = 9$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify

$25 (- b - 5) + 5 b + 9$ .

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Step 3.4.2.1.1

Simplify each term.

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Step 3.4.2.1.1.1

Apply the distributive property.

$4 = 25 (- b) + 25 \cdot - 5 + 5 b + 9$

$a = - b - 5$

$c = 9$

Step 3.4.2.1.1.2

Multiply

$- 1$ by

$25$ .

$4 = - 25 b + 25 \cdot - 5 + 5 b + 9$

$a = - b - 5$

$c = 9$

Step 3.4.2.1.1.3

Multiply

$25$ by

$- 5$ .

$4 = - 25 b - 125 + 5 b + 9$

$a = - b - 5$

$c = 9$

$4 = - 25 b - 125 + 5 b + 9$

$a = - b - 5$

$c = 9$

Step 3.4.2.1.2

Simplify by adding terms.

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Step 3.4.2.1.2.1

Add

$- 25 b$ and

$5 b$ .

$4 = - 20 b - 125 + 9$

$a = - b - 5$

$c = 9$

Step 3.4.2.1.2.2

Add

$- 125$ and

$9$ .

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

$4 = - 20 b - 116$

$a = - b - 5$

$c = 9$

Step 3.5

Solve for

$b$ in

$4 = - 20 b - 116$ .

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Step 3.5.1

Rewrite the equation as

$- 20 b - 116 = 4$ .

$- 20 b - 116 = 4$

$a = - b - 5$

$c = 9$

Step 3.5.2

Move all terms not containing

$b$ to the right side of the equation.

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Step 3.5.2.1

Add

$116$ to both sides of the equation.

$- 20 b = 4 + 116$

$a = - b - 5$

$c = 9$

Step 3.5.2.2

Add

$4$ and

$116$ .

$- 20 b = 120$

$a = - b - 5$

$c = 9$

$- 20 b = 120$

$a = - b - 5$

$c = 9$

Step 3.5.3

Divide each term in

$- 20 b = 120$ by

$- 20$ and simplify.

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Step 3.5.3.1

Divide each term in

$- 20 b = 120$ by

$- 20$ .

$\frac{- 20 b}{- 20} = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Step 3.5.3.2

Simplify the left side.

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Step 3.5.3.2.1

Cancel the common factor of

$- 20$ .

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Step 3.5.3.2.1.1

Cancel the common factor.

$\frac{- 20 b}{- 20} = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Step 3.5.3.2.1.2

Divide

$b$ by

$1$ .

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

$b = \frac{120}{- 20}$

$a = - b - 5$

$c = 9$

Step 3.5.3.3

Simplify the right side.

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Step 3.5.3.3.1

Divide

$120$ by

$- 20$ .

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

$b = - 6$

$a = - b - 5$

$c = 9$

Step 3.6

Replace all occurrences of

$b$ with

$- 6$ in each equation.

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Step 3.6.1

Replace all occurrences of

$b$ in

$a = - b - 5$ with

$- 6$ .

$a = - (- 6) - 5$

$b = - 6$

$c = 9$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify

$- (- 6) - 5$ .

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Step 3.6.2.1.1

Multiply

$- 1$ by

$- 6$ .

$a = 6 - 5$

$b = - 6$

$c = 9$

Step 3.6.2.1.2

Subtract

$5$ from

$6$ .

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

$a = 1$

$b = - 6$

$c = 9$

Step 3.7

List all of the solutions.

$a = 1, b = - 6, c = 9$

Step 4

Substitute the actual values of

$a$ ,

$b$ , and

$c$ into the formula for a quadratic equation to find the resulting equation.

$y = x^{2} - 6 x + 9$

Step 5

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