Consider the quadratic function f(x) = 1/5x2 – 5x + 12. Which statements are true about the function and its graph? Select three options. The value of f(–10) = 82 The graph of the function is a parabola. The graph of the function opens down. The graph contains the point (20, –8). The graph contains the point (0, 0).

Respuesta :

Answer:

The function f(x) = 1/5x2 – 5x + 12 is a quadratic function, because the higher exponent is 2.

The correct statements are:

  • The value of f(-10) = 82
  • The graph of the function is a parabola
  • The graph contains the point (20,-8)

Step-by-step explanation:

  • The value of f(-10) = 1/5(-10)2 – 5(-10) + 12 = 1/5(100)+50+12 = 82 (True statement).
  • The graph of the function is a parabola, as all the quadratic functions. (True statement).
  • The graph of the function opens down: The parabola opens up or down depending on the "a". It is the important part of the function  that determines whether it opens up or down, the parabola will open up when "a" is greater than zero. When "a" is smaller than zero the parabola will open down. In our example a = 1/5 (greater than zero, so the parabola opens up). (False Statement)
  • The graph contains the point (20,-8). For f(20) = 1/5(20)2 - 5(20) +12 = 1/5(400) - 100 + 12 = 80 - 100 + 12 = -8 (True Statement).
  • The graph contains the point (0,0). For f(0) = 1/5(0)2 - 5(0) + 12 = 12, the correct point is (0,12) (False Statement).

The given problem can be solved by using the properties of parabola.

The statements are true about the function and its graph are as follows;

The value of the function when f(-10) is 82.

The value of a is 1/5 which is positive then the parabola opens upward.

The graph contains the point (20, –8)

Given that,

Consider the quadratic function;

[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12.[/tex]

We have to determine,

Which statements are true about the function and its graph?

According to the question,

The quadratic function;

[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12.[/tex]

1. The value of the function when f(-10) is,

[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12\\\\f(-10) = \dfrac{1}{5}(-10)^2 -5(-10)+ 12\\\\ f(-10) = \dfrac{1}{5}(100) +50+ 12\\\\ f(-10) =20 +62\\\\ f(-10) = 82[/tex]

The value of the function when f(-10) is 82.

2. The graph of the function opens down

.[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12.[/tex]

The value of a is 1/5 which is positive then the parabola opens upward.

3.  The graph contains the point f(x) = 20,

[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12\\\\f(20) =\dfrac{1}{5}(20)^2 -5(20) + 12\\\\f(20) =\dfrac{1}{5}400 -100 + 12\\\\f(20) =80 -100 + 12\\\\f(20)= -20+12\\\\f(20)=-8[/tex]

The graph contains the point (20, –8)

4. The graph contains the point f(x) = 0,

[tex]\rm f(x) =\dfrac{1}{5}x^2 -5x + 12\\\\f(0) =\dfrac{1}{5}(0)^2 -5(0) + 12\\\\f(0) =\dfrac{1}{5}0 -0 + 12\\\\f(0) =0 -0 + 12\\\\f(0)= 12[/tex]

The graph contains the point (0, 12).

To know more about Parabola click the link given below.

https://brainly.com/question/4443998

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