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Match each set of points with the quadratic function whose graph passes through those points.

f(x) = x2 − 2x − 15
f(x) = -x2 − 2x + 15
f(x) = -x2 + 2x − 15

(0,-15), (1,-14), (2,-15)
(-2,15), (-1,16), (0,15)
(-3,0), (0,-15), (5,0)

1 Answer

3 votes

Answer:

f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

Explanation:

* Lets explain how to solve this question

- To find the points whose graph passes through them substitute the

x-coordinate in the function if the answer is the same with the

y-coordinate of the point then the graph passes through this point

lets do that

- Check the first set of points with the first function

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (1 , -14)

∵ f(x) = x² - 2x - 15

∴ f(0) = (1)² - 2(1) - 15 = -16 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (1 , -14)

∴ The graph does not pass through this set of points

- Check the second set of points with the first function

# Pint (-2 , 15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (-2)² - 2(-2) - 15 = 4 + 4 - 15 -7 ⇒ not same value of y-coordinate

∴ The graph of the function does not pass through point (-2 , 15)

∴ The graph does not pass through this set of points

- Check the third set of points with the first function

# Pint (-3 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (-3)² - 2(-3) - 15 = 9 + 6 -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-3 , 0)

# Pint (0 , -15)

∵ f(x) = x² - 2x - 15

∴ f(0) = (0)² - 2(0) - 15 = -15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , -15)

# Pint (5 , 0)

∵ f(x) = x² + 2x - 15

∴ f(0) = (5)² - 2(5) - 15 = 25 - 10 -15 = 0 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (5 , 0)

∴ The graph passes through this set of points

* f(x) = x² - 2x - 15 passes through (-3 , 0) , (0 , -15) , (5 , 0)

- Check the first set of points with the second function

# Pint (0 , -15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ not same value of y-coordinate

∴ The graph of the function does not passes through point (0 , -15)

∴ The graph does not pass through this set of points

- Check the second set of points with the second function

# Pint (-2 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-2)² - 2(-2) + 15 = -4 + 4 + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-2 , 15)

# Pint (-1 , 16)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(-1)² - 2(-1) + 15 = -1 + 2 + 15 = 16 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (-1 , 16)

# Pint (0 , 15)

∵ f(x) = -x² - 2x + 15

∴ f(0) = -(0)² - 2(0) + 15 = 15 ⇒ same value of y-coordinate

∴ The graph of the function passes through point (0 , 15)

The graph passes through this set of points

* f(x) = -x² - 2x + 15 passes through (-2 , 15) , (-1 , 16) , (0 , 15)

- Now we have the first set of points and the third function

The graph passes through this set of points

∴ f(x) = -x² + 2x - 15 passes through (0 , -15) , (1 , -14) , (2 , -15)

User Rezwan
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