Question

the function ( y = 20x^{2}+12x - 8 ) is graphed in the standard ( (x,y) ) coordinate plane. through which of the following intervals is the function increasing only?
a ( (1,3) )
c ( -1,3 )

Explanation:

Step1: Find the derivative of the function

The function is \(y = 20x^{2}+12x - 8\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), the derivative \(y^\prime=40x + 12\).

Step2: Find the critical point

Set \(y^\prime = 0\), so \(40x+12 = 0\). Solving for \(x\):

$$
LATEXBLOCK0
$$

For a quadratic function \(y = ax^{2}+bx + c\) (\(a = 20>0\)), the function is increasing when \(y^\prime>0\). Solving \(40x + 12>0\) gives \(x>-\frac{3}{10}\).

Step3: Check each interval

• For interval \(A=(1,3)\):

Since \(1>-\frac{3}{10}\), for all \(x\in(1,3)\), \(y^\prime = 40x + 12>0\). The function is increasing on \((1,3)\).

• For interval \(C=[-1,3]\):

When \(x=-1\), \(y^\prime=40\times(-1)+12=-28<0\). The function is not increasing at \(x = - 1\) in the interval \([-1,3]\).

Answer:

A. \((1,3)\)