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andreyandreev [35.5K]
3 years ago
10

Which graph represents a linear function? A coordinate plane is shown with a way line beginning along the x axis and continuing

along the graph. It curves up and down reaching y equals 1 and y equals negative 1 in a steady pattern along the horizontal axis A coordinate plane is shown with a parabola opening in a downward U shape. The vertex of the parabola is at y equals 7 point 5 and opening downward on either side of the y-axis from there A coordinate plane is shown with a line crossing through the y axis at negative 3 and the x axis at 2. A coordinate plane is shown with a line that begins to the left of the y axis passing through negative 1 comma negative 4, then curving right and passing through the y-axis at negative 4. It curves again and continues upward, passing through the x axis at around x equals 1.
Mathematics
2 answers:
Sliva [168]3 years ago
5 0
<span><span>-2-2 + 2 = 0(-2, 0)</span><span>-1-1 + 2 = 1(-1, 1)</span><span>00 + 2 = 2(0, 2)</span><span>11 + 2 = 3(1, 3)</span><span>22 + 2 = 4<span>(2, 4)</span></span></span>
Assoli18 [71]3 years ago
4 0

Answer:

A coordinate plane is shown with a line crossing through the y axis at negative 3 and the x axis at 2.

Step-by-step explanation:

The graph of a linear function is a straight line. So, we have to find the one graph that describes a straight line.

A coordinate plane is shown with a way line beginning along the x axis and continuing along the graph. It curves up and down reaching y equals 1 and y equals negative 1 in a steady pattern along the horizontal axis. This answer is incorrect because it does not describe a straight line, the line curves up and down between y = 1 and y = -1.

A coordinate plane is shown with a parabola opening in a downward U shape. The vertex of the parabola is at y equals 7 point 5 and opening downward on either side of the y-axis from there. This answer is incorrect because describes a parabola which is the graph of a quadratic function.

A coordinate plane is shown with a line crossing through the y axis at negative 3 and the x axis at 2. This is the correct answer because describes a straight line that passes through the points (0 , - 3) and (2 , 0).

A coordinate plane is shown with a line that begins to the left of the y axis passing through negative 1 comma negative 4, then curving right and passing through the y-axis at negative 4. It curves again and continues upward, passing through the x axis at around x equals 1. This answer is incorrect because it does not describe a straight line.

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Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

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The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = 3^{x + 2}

The asymptote is given as x → -∞, f(x) = 3^{x + 2} → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = 3^{0 + 2} = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = 5^{1  - x}

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = 5^{1  - 0} = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = 2^{x + 3} - 1

The asymptote is 2^{x + 3}  → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) =  2^{0 + 3} - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, 2^{x + 3} - 1 = 0

2^{x + 3} = 1

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The x-intercept is (-3, 0)

f) f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2

The asymptote is \left (\dfrac{1}{2} \right)^{x - 2} → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6

The y-intercept is (0, 6)

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