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IceJOKER [234]
2 years ago
14

Given the definitions of f(x)f(x) and g(x)g(x) below, find the value of (f\circ g)(-4).(f∘g)(−4). f(x)= f(x)= \,\,3x^2+4x-3 3x 2

+4x−3 g(x)= g(x)= \,\,-2x-11 −2x−11
Mathematics
1 answer:
rewona [7]2 years ago
8 0

Answer:

(f∘g)(−4) = 12

Step-by-step explanation:

Given the functions

f(x)= 3x^2 + 4x - 3

g(x) = -2x-11

Required

(f∘g)(−4)

(f∘g)(x) = f(g(x))

f(g(x)) = f(-2x-11)

f(-2x-11) = 3(-2x-11)^2 + 4(-2x-11) - 3

f(g(x)) = 3(-2x-11)^2 + 4(-2x-11) - 3

f(g(-4)) = 3(-2(-4)-11)^2 + 4(-2(-4)-11) -3

f(g(-4)) = 3(8-11)^2 + 4(8-11) - 3

f(g(-4)) = 3(-3)^2 + 4(-3) - 3

f(g(-4))  = 3(9) - 12 - 3

f(g(-4))  = 27 - 15

f(g(-4)) = 12

Hence (f∘g)(−4) = 12

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              \bold{2.\quad x=-\dfrac{1}{2}(y-10)^2}+1

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<u>Step-by-step explanation:</u>

Notes: Vertex form is: y =a(x - h)² + k    or      x =a(y - k)² + h

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  • point of vertex is midpoint of focus and directrix:   \dfrac{focus+directrix}{2}

     \bullet\quad a=\dfrac{1}{4p}

  • p is the distance from the vertex to the focus

1)

focus = \bigg(\dfrac{-31}{4},2\bigg)\qquad directrix: x=\dfrac{-33}{4}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{-31}{4}+\frac{-33}{4}}{2}=\dfrac{\frac{-64}{4}}{2}=\dfrac{-16}{2}=-8\\\\\text{The y-value of the vertex is given by the focus as: 2}\\\\\text{vertex (h, k)}=(-8,2)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{-31}{4}-\dfrac{-32}{4}=\dfrac{1}{4}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{4})}=\dfrac{1}{1}=1

Now, plug in a = 1   and    (h, k) = (-8, 2) into the equation x =a(y - k)² + h

x = (y - 2)² + 8

***************************************************************************************

2)

focus = \bigg(\dfrac{1}{2},10\bigg)\qquad directrix: x=\dfrac{3}{2}\\\\\text{Since directrix is x, then the x-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{1}{2}+\frac{3}{2}}{2}=\dfrac{\frac{4}{2}}{2}=\dfrac{2}{2}=1\\\\\text{The y-value of the vertex is given by the focus as: 10}\\\\\text{vertex (h, k)}=(1,10)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{1}{2}-\dfrac{2}{2}=\dfrac{-1}{2}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{-1}{2})}=\dfrac{1}{-2}=-\dfrac{1}{2}

Now, plug in a = -1/2   and    (h, k) = (1, 10) into the equation x =a(y - k)² + h

\bold{x=-\dfrac{1}{2}(y-10)^2}+1

***************************************************************************************

3)

focus = \bigg(-9,\dfrac{57}{8}\bigg)\qquad directrix: y=\dfrac{55}{8}\\\\\text{Since directrix is y, then the y-value of the vertex is:}\\\\\dfrac{focus+directrix}{2}=\dfrac{\frac{57}{8}+\frac{55}{8}}{2}=\dfrac{\frac{112}{8}}{2}=\dfrac{14}{2}=7\\\\\text{The x-value of the vertex is given by the focus as: -9}\\\\\text{vertex (h, k)}=(-9,7)

Now let's find the a-value:

p=focus-vertex\\\\p=\dfrac{57}{8}-\dfrac{56}{8}=\dfrac{1}{8}\\\\\\a=\dfrac{1}{4p}=\dfrac{1}{4(\frac{1}{8})}=\dfrac{1}{\frac{1}{2}}=2

Now, plug in a = 2   and    (h, k) = (-9, 7) into the equation y =a(x - h)² + k

y = 2(x +9)² + 7

4 0
3 years ago
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