consider the quadratic function $f(x)=\frac{1}{5}x^{2}-5x + 12$. which statements are true about the…

consider the quadratic function $f(x)=\frac{1}{5}x^{2}-5x + 12$. which statements are true about the function and its graph? select three options.\nthe value of $f(-10)=82$\nthe graph of the function is a parabola.\nthe graph of the function opens down.\nthe graph contains the point $(20,-8)$.\nthe graph contains the point $(0,0)$.

consider the quadratic function $f(x)=\frac{1}{5}x^{2}-5x + 12$. which statements are true about the function and its graph? select three options.\nthe value of $f(-10)=82$\nthe graph of the function is a parabola.\nthe graph of the function opens down.\nthe graph contains the point $(20,-8)$.\nthe graph contains the point $(0,0)$.

Answer

Answer:

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

Explanation:

Step1: Calculate (f(-10))

Substitute (x = - 10) into (f(x)=\frac{1}{5}x^{2}-5x + 12). [ \begin{align*} f(-10)&=\frac{1}{5}\times(-10)^{2}-5\times(-10)+12\ &=\frac{1}{5}\times100 + 50+12\ &=20 + 50+12\ &=82 \end{align*} ]

Step2: Identify the shape of the graph

The general form of a quadratic function is (y = ax^{2}+bx + c), and its graph is a parabola. Here (a=\frac{1}{5}), (b=-5), (c = 12), so the graph of (f(x)) is a parabola.

Step3: Determine the direction of the parabola

For a quadratic function (y=ax^{2}+bx + c), if (a>0), the parabola opens up; if (a < 0), the parabola opens down. Since (a=\frac{1}{5}>0), the graph of the function opens up, so the statement "The graph of the function opens down" is false.

Step4: Check if the point ((20,-8)) is on the graph

Substitute (x = 20) into (f(x)=\frac{1}{5}x^{2}-5x + 12). [ \begin{align*} f(20)&=\frac{1}{5}\times20^{2}-5\times20+12\ &=\frac{1}{5}\times400-100 + 12\ &=80-100 + 12\ &=-8 \end{align*} ] So the graph contains the point ((20,-8)).

Step5: Check if the point ((0,0)) is on the graph

Substitute (x = 0) into (f(x)=\frac{1}{5}x^{2}-5x + 12). [ \begin{align*} f(0)&=\frac{1}{5}\times0^{2}-5\times0+12\ &=12\neq0 \end{align*} ] So the graph does not contain the point ((0,0)).