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natulia [17]
3 years ago
13

Consider the linear function f(x)=4x-5 Find the values of x for which f(x)=3,f(x)=5,f(x)=-3

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

f(2) = 3

f(5/2) = 5

f(½) = -3

Step-by-step explanation:

Given the linear function f(x) = 4x - 5:

<h3>f(x) = 3</h3>

3 = 4x - 5

3 + 5 = 4x - 5 + 5

8 = 4x

8/4 = 4x/4

2 = x

Therefore, when f(x) = 3, x = 2.

<h3>f(x) = 5</h3>

5 = 4x - 5

5 + 5 = 4x - 5 + 5

10 = 4x

10/4 = 4x/4

5/2 = x

Therefore, when f(x) = 5, x = 5/2.

<h3>f(x) = -3</h3>

-3 = 4x - 5

-3 + 5 = 4x - 5 + 5

2 = 4x

2/4 = 4x/4

½ = x

Therefore, when f(x) = -3, x = ½.

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Answer:

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Step-by-step explanation:

1st boat:

Parabola equation:

y=ax^2 +bx+c

The x-coordinate of the vertex:

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Equation:

y=ax^2 -2ax+c

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1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1

Solve:

c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}

Parabola equation:

y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}

2nd boat:

Parabola equation:

y=ax^2 +bx+c

The x-coordinate of the vertex:

x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0

Equation:

y=ax^2+c

The y-coordinate of the vertex:

y_v=a\cdot 0^2+c\Rightarrow c=-7

Parabola passes through the point (-8,1), so

1=a\cdot (-8)^2-7\\ \\64a-7=1

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Parabola equation:

y=\dfrac{1}{8}x^2 -7

System of two equations:

\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.

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