Question

Please help! just need answers for a and b, please!!!!

Answer 1

see explanation

Explanation:

(a)

A lies on the y- axis with x- coordinate of zero

substitute x = 0 into either of f(x) or g(x) for y- coordinate

g(0) = 2(0) + 1 = 0 + 1 = 1

then A (0, 1 )

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B and C lie on the x- axis with y- coordinate of zero

let f(x) = 0 and solve for x , that is

1 - x² = 0 , then

x² = 1 ( take square root of both sides )

x = ±
`√(1)` = ± 1

then B (- 1, 0 ) and C (1, 0 )

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D is the point on the x- axis where g(x) crosses

let g(x) = 0 and solve for x

2x + 1 = 0 ( subtract 1 from both sides )

2x = - 1 ( divide both sides by 2 )

x = -
`(1)/(2)`

then D (-
`(1)/(2)` , 0 )

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(b)

E is the point of intersection of f(x) and g(x) so equate them, that is

2x + 1 = 1 - x² ( subtract 1 - x² from both sides )

x² + 2x = 0 ← factor out x from each term

x(x + 2) = 0

equate each factor to zero and solve for x

x = 0 ← x- coordinate of A

x + 2 = 0 ⇒ x = - 2 ← x- coordinate of E

substitute x = - 2 into g(x) for corresponding y- coordinate

g(- 2) = 2(- 2) + 1 = - 4 + 1 = - 3

then E (- 2, - 3 )

Answer 2

The answers are :

a)

• A = (0, 1)
• B = (-1, 0)
• C = (1, 0)
• D = (-0.5, 0)

b)

• E = (-2, -3)

Let's find point A by using g(x). A is clearly the y-intercept so take x = 0.

• g(0) = 2(0) + 1
• g(0) = 1
• A = (0, 1)

Points B and C can be found using f(x). As they are the x-intercepts, take y = 0.

• 0 = 1 - x²
• x² = 1
• x = ±1

As B lies to the left of the y-axis, its x-coordinate is negative.

• B = (-1, 0)

As C lies to the right of the y-axis, its x-coordinate is positive.

• C = (1, 0)

Point D can be found using g(x). D is clearly the x-intercept so take y = 0.

• 0 = 2x + 1
• 2x = -1
• x = -1/2 = -0.5
• D = (-0.5, 0)

Point E lies at the intersection of f(x) and g(x). So they have to be equated.

• f(x) = g(x)
• 1 - x² = 2x + 1
• x² + 2x = 0
• x (x + 2) = 0
• x = -2
• g(-2) = 2(-2) + 1
• g(-2) = -4 + 1
• g(-2) = -3
• E = (-2, -3)