Question

6.
which equation could be used to find the zeros of the function
$f(x) = 3x^2 - 14x - 5$
(a) $(3x + 1)(x + 5) = 0$
(b) $(3x + 1)(x - 5) = 0$
(c) $(3x - 1)(x + 5) = 0$
(d) $(x + 1)(3x - 5) = 0$
objective:n.q.1 use units as a way to understand problems and to guide the solution of multi - step problems; choose
and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays.

1. a graph represents the distance traveled by a car over a period of time. the y - axis

is labeled \distance (miles)\ and the x - axis is labeled \time (minutes)\.
which unit would be appropriate for the rate of change in the graph?
a. miles per hour
b. minutes per mile
c. miles per minute
d. hours per mile

1. a graph represents the volume of water flowing out of a tank over a period of time.

the y - axis is labeled \volume (gallons)\ and the x - axis is labeled \time (seconds)\.
which unit would be appropriate for the rate of change in the graph?
(a) seconds per gallon
(b) gallons per second

Explanation:

First Question (Finding Zeros of Quadratic Function)

Step1: Recall factoring method

To find the zeros of \( f(x) = 3x^2 - 14x - 5 \), we factor the quadratic. We need two numbers that multiply to \( 3\times(-5)= -15 \) and add to \( -14 \). The numbers are \( -15 \) and \( 1 \).

Step2: Rewrite and factor

Rewrite the middle term: \( 3x^2 - 15x + x - 5 \). Group: \( (3x^2 - 15x) + (x - 5) = 3x(x - 5) + 1(x - 5) = (3x + 1)(x - 5) \). Set to zero: \( (3x + 1)(x - 5) = 0 \).

Step1: Understand rate of change

Rate of change is \( \frac{\text{Change in } y}{\text{Change in } x} \). Here, \( y \)-axis is distance (miles), \( x \)-axis is time (minutes). So rate is \( \frac{\text{miles}}{\text{minutes}} \), i.e., miles per minute.

Step2: Evaluate options

Option A: miles per hour (time in hours, not minutes). Option B: minutes per mile (inverse). Option C: miles per minute (matches \( \frac{\text{miles}}{\text{minutes}} \)). Option D: hours per mile (inverse and wrong time unit).

Step1: Recall rate of change formula

Rate of change is \( \frac{\text{Change in } y}{\text{Change in } x} \). \( y \)-axis: volume (gallons), \( x \)-axis: time (seconds). So rate is \( \frac{\text{gallons}}{\text{seconds}} \), i.e., gallons per second.

Step2: Evaluate options

Option (a): seconds per gallon (inverse). Option (b): gallons per second (matches \( \frac{\text{gallons}}{\text{seconds}} \)).

Answer:

(b) \( (3x + 1)(x - 5) = 0 \)

Second Question (Rate of Change for Car Distance)