Online Course Discussion Forum
Math Challenge I-C Handout 6
Please help me with the following questions in handout 6's assignment.
Question 6.21: I'm not familiar with how the exponent tricks work even after watching the lecture. Can I please have a detailed explanation? I'm kind of slow at getting things. :p
Part a):
Part b):
Question 6.30: For this problem, I tried plugging in x and y values into the equation, but the answer to each pair is different.
y−1=a(x−10)^2
if x=2
6.25=a(2-10)^2
6.25=64a
a~0.1
However: When I try a different set, I get a different a value. HELP!!
For 6.21, remember that negative exponents mean flipping the fraction upside down. For example $$2^{-2} = \left(\frac{2}{1}\right)^{-2} = \left(\frac{1}{2}\right)^2 \text{ or } \left(\frac{2}{3}\right)^{-3.5} = \left(\frac{3}{2}\right)^{3.5}.$$
Also, fractions mean square, cube, fourth, etc. roots. So $2^{1/2} = \sqrt{2}$ or $16^{1/4} = \sqrt[4]{16}$ or $5^{3/2} = (5^{1/2})^3 = (\sqrt{5})^3$. For part b), try to write both the numerator and denominator as $2^{\text{something}}$ first and then use the fact that $x^a / x^b = x^{a-b}$.
For 6.30, remember that when finding a model there could be different possible values. One natural way to proceed would be using a similar point that you used in 6.29 (where you probably estimated the $y$ value when $x = 0$).
So i get the 6.21 parts, but the modeling is still confusing. I plugged in the y value when x=0 into the equation, but the answer is 0 which i'm pretty sure is not the one you are looking for.
Looking at the points in the graph, an estimate might be that when $x = 0$, then $y = 10$. Using this you can solve for $a$. Note using this number you'll still need to "approximate" the value of $a$ so that it is of the form $1/K$ for an integer $K$.
In the example you gave you had $a \approx 0.1 \approx \frac{1}{10}$ (so $K=10$). This number is fairly close to the one you'll get using the point $(0,10)$.
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