Question

a scientist is measuring the temperature of a chemical undergoing a reaction during an experiment. they found th
the temperature of the chemical, y, can be expressed as a quadratic function of time, x.
this quadratic function can be written in three different forms, all shown below. they are written in terms of x
(seconds after the experiment began) and y (temperature in degrees fahrenheit).
which of these forms is most useful for the scientist to be able to determine the chemicals temperature whenever
the experiment first begins?

show your work here

y = 4x² - 32x + 60\t\t\t\t
y = 4(x - 3)(x - 5)

y = 4(x - 4)² - 4

Explanation:

Step1: Understand the problem

The experiment begins at \( x = 0 \) (time in seconds). We need to find which quadratic form is easiest to evaluate at \( x = 0 \) to get the initial temperature \( y \).

Step2: Analyze each form at \( x = 0 \)

• For \( y = 4x^2 - 32x + 60 \):

Substitute \( x = 0 \): \( y = 4(0)^2 - 32(0) + 60 = 60 \). This is straightforward as we just plug in \( x = 0 \), and the constant term is the value at \( x = 0 \).

• For \( y = 4(x - 3)(x - 5) \):

Substitute \( x = 0 \): \( y = 4(0 - 3)(0 - 5)=4(-3)(-5)=60 \). But we need to multiply two terms first, which is more work than the standard form here.

• For \( y = 4(x - 4)^2 - 4 \):

Substitute \( x = 0 \): \( y = 4(0 - 4)^2 - 4 = 4(16)-4 = 64 - 4 = 60 \). This requires squaring a term and then multiplying, more steps than the standard form.

The standard form \( y = 4x^2 - 32x + 60 \) allows us to directly see that when \( x = 0 \), the first two terms are zero, and we just use the constant term. So it's the most useful for finding the initial temperature.

Answer:

The form \( y = 4x^2 - 32x + 60 \) (the first option) is most useful.