questions c, d, e please!

Answer:
c) 3 units
d) g(x) - f(x) = x² + 2x
e) (-∞, -2] ∪ [0, ∞)
Step-by-step explanation:
To calculate the length of FC, first find the coordinates of point C.
The y-value of point C is zero since this is where the function f(x) intercepts the x-axis. Therefore, set f(x) to zero and solve for x:
[tex]\implies 1-x^2=0[/tex]
[tex]\implies x^2=1[/tex]
[tex]\implies \sqrt{x^2}=\sqrt{1}[/tex]
[tex]\implies x= \pm 1[/tex]
As point C has a positive x-value, C = (1, 0).
To find point F, substitute the x-value of point C into g(x):
[tex]\implies g(1)=2(1)+1=3[/tex]
⇒ F = (1, 3).
Length FC is the difference in the y-value of points C and F:
[tex]\begin{aligned} \implies \sf FC& = \sf y_F-y_C\\ & = \sf 3-0\\ & =\sf 3\:units \end{aligned}[/tex]
Given functions:
[tex]\begin{cases}f(x)=1-x^2\\ g(x)=2x+1 \end{cases}[/tex]
Therefore:
[tex]\begin{aligned}\implies g(x)-f(x) & = (2x+1) - (1-x^2)\\& = 2x+1-1+x^2\\& = x^2+2x\end{aligned}[/tex]
The values of x for which g(x) ≥ f(x) are where the line of g(x) is above the curve of f(x):
Point A is the y-intercept of both functions, therefore the x-value of point A is 0.
To find the x-value of point E, equate the two functions and solve for x:
[tex]\begin{aligned}g(x) & = f(x)\\\implies 2x+1 & = 1-x^2\\x^2+2x & = 0\\x(x+2) & = 0\\\implies x & = 0, -2\end{aligned}[/tex]
As the x-value of point E is negative ⇒ x = -2.
Therefore, the values of x for which g(x) ≥ f(x) are:
Answer:
a)
A = (0, 1)
B = (-1, 0)
C = (1, 0)
D = (-0.5, 0)
b) E = (-2, -3)
c) FC = 3 units
d) x² + 2x
e) x ≤ -2 and x ≥ 0
Explanation:
This question displays one equation of a linear function g(x) = 2x + 1 and a parabolic function f(x) = 1 - x².
a)
A point is where the linear function cuts the y axis.
y = 1 - (0)²
y = 1
A = (0, 1)
B and C point is where the parabolic function cuts the x axis.
1 - x² = 0
-x² = -1
x² = 1
x = ±√1
x = -1, 1
B = (-1, 0), C = (1, 0)
D point is where the linear function cuts x axis.
2x + 1 = 0
2x = -1
x = -1/2 or -0.5
D = (-0.5, 0)
b)
E point is where both equations intersect each other.
y = y
2x + 1 = 1 - x²
x² + 2x = 0
x(x + 2) = 0
x = 0, x = -2
y = 1, y = -3
E = (-2, -3)
c)
C : (1, 0)
To find F point
y = 2(1) + 1
y = 3
F : (1, 3)
[tex]\sf Distance \ between \ two \ points = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
[tex]\sf d = \sqrt{(1 - 1)^2 + (3 - 0)^2}[/tex]
[tex]\sf d = \sqrt{0 + 3^2}[/tex]
[tex]\sf d = 3[/tex]
FC length = 3 units
d)
g(x) - f(x)
(2x + 1) - (1 - x²)
2x + 1 - 1 + x²
x² + 2x
e)
g(x) ≥ f(x)
2x + 1 ≥ 1 - x²
x² + 2x ≥ 0
x(x + 2) ≥ 0
[tex]\boxed{If \ x \ \geq \ \pm \ a \ then \ -a \ \leq x \ \ and \ x \ \geq \ a }[/tex]
x ≤ -2 and x ≥ 0