Help? So confused.

Question

1.Create a graph of the pH function either by hand or using technology. Locate on your graph where the pH value is 0 and where it is 1. You may need to zoom in on your graph.

2.The pool maintenance man forgot to bring his logarithmic charts, and he needs to raise the amount of hydronium ions, t, in the pool by 0.50. To do this, he can use the graph you created. Use your graph to find the pH level if the amount of hydronium ions is raised to 0.50. Then, convert the logarithmic function into an exponential function using y for the pH.

3. The pool company developed new chemicals that transform the pH scale. Using the pH function p(t) = −log10t as the parent function, explain which transformation results in a y-intercept and why. You may graph by hand or using technology. Use complete sentences and show all translations on your graph.
p(t) + 1
p(t + 1)
−1 • p(t)

Answer 1

There is a lot going on in this one, so let's take the three parts in order. Everything comes from the same parent function:

p(t) = −log₁₀t

For the first part, the two points they are asking you to locate are easier to find algebraically than by zooming around on the graph. Where the pH is 0 we set −log₁₀t = 0, which gives log₁₀t = 0, and since 10⁰ = 1 that means t = 1. Where the pH is 1 we set −log₁₀t = 1, so log₁₀t = −1, and since 10⁻¹ = 0.1 that means t = 0.1. So your curve passes through (1, 0) and (0.1, 1). Notice the curve comes down from the left and crosses the horizontal axis at t = 1, and that there is a vertical asymptote at t = 0 because you cannot take the log of zero.

For the second part, put t = 0.50 into the function:

p(0.50) = −log₁₀(0.50) = log₁₀2 ≈ 0.30

so the pH would be about 0.30. To turn the logarithmic form into exponential form, remember that a log and an exponent are just two ways of saying the same sentence. Starting from y = −log₁₀t, multiply both sides by −1 to get log₁₀t = −y, and then read that as an exponent statement:

t = 10^−y

You can check it with the number we just found, since 10^−0.30 comes out to about 0.50.

The third part is the interesting one, and the key is to think about the domain instead of the shape. A y-intercept is just the point where t = 0, so the question is really asking which of these transformations lets t = 0 be a legal input. The parent function has a vertical asymptote at t = 0, so t = 0 is not allowed and there is no y-intercept to begin with.

Adding 1 on the outside, p(t) + 1, slides the whole graph up one unit. Sliding up does nothing to what you are allowed to put in, so the asymptote stays at t = 0 and there is still no y-intercept. Multiplying by −1, so −1 · p(t) = log₁₀t, flips the graph over the horizontal axis, and again that does not change the inputs, so the asymptote stays at t = 0 and there is still no y-intercept.

But p(t + 1) = −log₁₀(t + 1) shifts the graph one unit to the LEFT, and that drags the asymptote along with it from t = 0 to t = −1. Now t = 0 is inside the domain, and putting it in gives −log₁₀(0 + 1) = −log₁₀1 = 0.

So, the transformation that produces a y-intercept is p(t + 1), and the intercept is at (0, 0). The reason is that it is the only one of the three that moves the function horizontally, and a horizontal shift is the only thing that changes which inputs are allowed.