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)

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)

Step-by-step 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)

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.