Derive the equation of the parabola with a focus at (−2, 4) and a directrix of y = 6. put the equation in standard form.

f(x) = one fourthx2 − x + 4
f(x) = −one fourthx2 − x + 4
f(x) = one fourthx2 − x + 5
f(x) = −one fourthx2 − x + 5

B

Step-by-step explanation:

From any point (x, y) on the parabola the focus and directrix are equidistant.

Using the distance formula

= | y - 6 |

Squaring both sides

(x + 2)² + (y - 4)² = (y - 6)² ←  distributing

x² + 4x + 4 + y² - 8y + 16 = y² - 12y + 36 ( subtract y² - 12y + 36 from both sides )

x² + 4x + 4 + 4y - 20 = 0 ( subtract x² + 4x + 4 from both sides )

4y - 20 = - x² - 4x - 4 ( add 20 to both sides )

4y = - x² - 4x + 16 ( divide through by 4 )

y = - x² - x + 4, that is

f(x) = - x² - x + 4 → B

 See below

Step-by-step explanation:

4) You made an error in step 3

Step 1: -x³ + 22x² - 121x = 0

Step 2: -x(x² - 22x + 121) = 0

Step 3: -x(x - 11²) = 0       Error!

           -x(x - 11)² = 0        Correction

Step 4: -x = 0     (x - 11)² = 0

Step 5:    x = 0      x = 11

15) Your work is correct

    n = 8, n = -9      BOTH answers are correct

n = 8      Error!  

              n = 8, n = -9    Correction

18) The formula is: h(t) =at² + vt + h(0)

h(t) = -4.9t² + 24.5t + 117.6      option 3

19) Now that you have the correct equation, you can use the quadratic formula to solve for t

t = -3, t = 8

Since time cannot be negative, t = -3 is not valid

t = 8    option 1

20) It is given that h(t) = 49

Equation: 49 = -4.9t² + 24.5t + 117.6

                 0 = -4.9t² + 24.5t + 68.6

Use the quadratic equation to solve: t = -2, t = 7

Since time cannot be negative, t = -2 is not valid

t =7