Answer:
see explanation
Step-by-step explanation:
A
given a parabola in standard form
f(x) = ax² + bx + c ( a ≠ 0 ) , then the x- coordinate of the vertex is
x = - [tex]\frac{b}{2a}[/tex]
f(x) = - 2x² - 4x + 14 ← is in standard form
with a = - 2, b = - 4 , then
x = - [tex]\frac{-4}{-4}[/tex] = - 1
substitute x = - 1 into f(x) for corresponding y- coordinate
f(- 1) = - 2(- 1)² - 4(- 1) + 14 = - 2 + 4 + 14 = 16
vertex = (- 1, 16 )
B
to find the y- intercept let x = 0
f(0) = - 2(0)² - 4(0) + 14 = 0 - 0 + 14
y- intercept = (0, 14 )
C
to find the x- intercepts let f(x) = 0 , that is
- 2x² - 4x + 14 = 0 ( divide through by - 2 )
x² + 2x - 7 = 0 ( add 7 to both sides )
x² + 2x = 7
using the method of completing the square
add ( half the coefficient of the x- term )² to both sides
x² + 2(1)x + 1 = 7 + 1
(x + 1)² = 8 ( take square root of both sides )
x + 1 = ± [tex]\sqrt{8}[/tex] ( subtract 1 from both sides )
x = - 1 ± [tex]\sqrt{8}[/tex]
then
x = - 1 - [tex]\sqrt{8}[/tex] ≈ - 3.83 ( to 2 dec. places )
x = - 1 + [tex]\sqrt{8}[/tex] ≈ 1.83 ( to 2 dec. places )
that is x- intercepts are (- 3.83, 0 ) , (1.83, 0 )
