Chapter 1.4, Problem 41E

Solution

To find: The zeros of the function $y=4x^2-19x-5$ by writing in intercept form.

  $x=-\frac{1}{4}$ , $x=5$ .

Given information:

The function, $y=4x^2-19x-5$ .

Concept used:

The intercept form of a quadratic function is given as $y=\left(x-a\right)\left(x-b\right)$ , where $a$ and $b$ are zeros of the function.

Calculation:

Rewrite the function in intercept form as follows:

  $\begin{array}{c}y=4x^2-19x-5\\ =\left(4x^2+x\right)+\left(-20x-5\right)\\ =x\left(4x+1\right)-5\left(4x+1\right)\\ y=\left(4x+1\right)\left(x-5\right)\end{array}$

The intercept of the function are the points where the factor $4x+1=0$ and $x-5=0$ .

Solve the two equations to find the intercepts.

  $\begin{array}{c}4x=-1\\ x=-\frac{1}{4}\end{array}$

And,

  $\begin{array}{c}x-5=0\\ x=5\end{array}$

Conclusion:

The zeros of the function are $x=-\frac{1}{4}$ and $x=5$ .