Question

write the equation of the parabola and a vertex form and standard form given the following information​

Answer 1

see explanation

Explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

(9)

Here (h, k) = (3, - 2), thus

y = a(x - 3)² - 2

To find a substitute (2, 3) into the equation

3 = a(2 - 3)² - 2 ( add 2 to both sides )

5 = a(- 1)² = a , then

y = 5(x - 3)² - 2 ← in vertex form

y = 5(x² - 6x + 9) - 2

= 5x² - 30x + 45 - 2

y = 5x² - 30x + 43 ← in standard form

(10)

Here (h, k) = (2, - 5), thus

y = a(x - 2)² - 5

To find a substitute (5, 4) into the equation

4 = a(5 - 2)² - 5 ( add 5 to both sides )

9 = a(- 3)² = 9a ( divide both sides by 9 )

a = 1

y = (x - 2)² - 5 ← in vertex form

y = x² - 4x + 4 - 5

y = x² - 4x - 1 ← in standard form

Answer 2

Answer: 9) y = 5(x - 3)² - 2 y = 5x² - 30x + 43

10) y = (x - 2)² - 5 y = x² - 4x + 2

Explanation:

The Vertex form of a quadratic equation is: y = a(x - h)² + k where

• "a" is the vertical stretch
• (h, k) is the vertex
• (x, y) is any point on the parabola

The Standard form of a quadratic equation is: y = ax² + bx + c

--> this is found by expanding the vertex form and adding like terms.

Input Vertex (h, k) and Point (x, y) into the Vertex form to solve for "a"

9) Vertex (h, k) = (3, -2) and Point (x, y) = (2, 3)

3 = a(2 - 3)² - 2

3 = a(-1)² - 2

5 = a(1)

5 = a

Vertex form: y = 5(x - 3)² - 2

Standard Form: y = 5(x² - 6x + 9) - 2

y = 5x² - 30x + 45 - 2

y = 5x² - 30x + 43

10) Vertex (h, k) = (2, -5) and Point (x, y) = (5, 4)

4 = a(5 - 2)² - 5

4 = a(3)² - 5

9 = a(9)

1 = a

Vertex form: y = (x - 2)² - 5

Standard Form: y = (x² - 4x + 4) - 2

y = x² - 4x + 2