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Course: Digital SAT Math > Unit 8
Lesson 11: Quadratic graphs: mediumManipulating quadratic and exponential expressions — Basic example
Watch Sal work through a basic Manipulating quadratic and exponential expressions problem.
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why make things so much harder if all you have to do is plug in for the exponent of 2. you get the same answer of .7225(37 votes)
Things were made simpler. That was 10 times smarter, and a 100 times faster.(22 votes)
- Why make things more complicated? This could have been answered with common sense, 85% of 85% is going to be less than .85 so you pick A because that is the only one that is less than .85 and move on.(21 votes)
Sometimes when your under a time crunch, you don't detect the simple tricks you could use and instead use methods you're more familiar with even if they're more complicated and time-consuming.(7 votes)
This isn't a question. But a quicker way is:
when the last digit of 2 different values add up to 10, and the first digit(s) are the same, you can simply multiply the first digit(s) by the next digit(s) in succession. Then multiply the last digits by one another and put it behind the former.
85 * 85 //5 + 5 = 10; both have the same digit in the tens place so
9 * 8 = 72
7225
24 * 26 = 624 // 6 + 4 = 10; 5 * 4 = 24
123 * 127 = 15621 // 12 * 13 = 156; 3 * 7 = 21(12 votes)
Wouldn't you just put 2 in place of t and solve that way?(5 votes)
can someone explain the definition of a constant? i don't understand what they are.(2 votes)
You can think of a constant as just a plain number. When talking polynomials, constants are everything that doesn't have any x-coefficients next to them. In the equation y = 5x + 3 for example, 3 would be your constant.
On a graph, if your equation is in the "y = " form then a constant added to the end of the equation would shift the graph up (if positive) or down (if negative) by that amont.(7 votes)
If you put 2 in place of the variable that means you do this:
V(t)=24900*0.85^2
V(t)=24900*0.7225
V(t)=17990.25
I am very confused, wouldn't it be be (D), because you also multiply by 24900.
Please help me out. I appreciate it.
Thank you(2 votes)
Okay so option D says 1.7 right?? which isn't correct in this case...You are right, we multiply by 24900 to get the value two years later, but we also have to divide it by the current value since the question says "how many times its present value" and the present value is 24900, so it is effectively (24900*(0.85^2))/24900 which is just 0.85^2=0.7225(7 votes)
I think that all you people are talking about another lesson, rather than talking about what Sal is explaining.
I would say there some issues with that video, maybe it mix up with another video.(5 votes)- Or...you could just multiply 0.85 by itself because after two years it will be 85% of 85%. It could be 30 seconds rather than 3 minutes.(4 votes)
You don't need to solve the question the answer is obvious by elimination(2 votes)- Yes, you could do that. But would you want to do that if this was not on the calculator allowed side of the SAT?(3 votes)
- Why didn't he chose option A it seemed true as well.(2 votes)
- You're correct that option A is an equivalent form of the equation in the graph, which is y = x^2 + x - 6. However, A) doesn't fit the second part of the question, which wants us to find an equation that has the x-intercepts as constants in the equation.
That means if the x-intercepts were x = 2 and -3 (like they are in the video), we would have to have 2 and -3 somewhere in our equation. Option D) accomplishes this because it has a 2 and a 3 in it, and is also equivalent to the equation of the graph, which C) isn't. Hope this helps!(2 votes)
Video transcript
- [Instructor] We're asked,
which of the following is an equivalent form of
the equation of the graph shown in the xy-plane from
which the x-intercepts can be identified as
constants in the equation? So pause this video and see
if you can figure this out. All right, now let's work
through this together. So we wanna find an equation
where if we graph it, we get this parabola right
over here, and we want it to be in the form in which or
from which the x-intercepts can be identified as
constants in the equation. Well, let's look at the x-intercepts here. We have an x-intercept right over here. X-intercepts happen
when y is equal to zero when we intersect the x axis. So one happens at x equals
two, and then another here happens at x is equal to negative three. And so when we look at the choices, we are looking for a two or a three and the first two really don't see that. You aren't able to pick out
these x-intercepts easily from choices A or B, so
you can rule those out. Now, both choices C and D have some things that deal with threes and twos here. And what we have to remember is these are the x values that make y equal to zero. And so when you have it
in this factored form, when you have this quadratic
here in a factored form, if you have x minus two,
that means that x equals two is going to be an x-intercept. How do we know that? Well, if you put a two in right over here, two minus two is zero, zero
times anything is zero. If you put in a negative three here, negative three plus three
is going to be zero. So this is actually the
choice that we are looking at. Once you factor this
quadratic in a form like this the intercepts are actually
going to be the opposite, the negatives, of these
numbers right over here. So this C right over here
was a distractor to say, Oh look, I see a negative
three and I see a two, but this is actually a different equation than what we have over here. And you would see here that
the x-intercepts would be at x equals three and
x equals negative two, not negative three and positive two, so we'd rule that one out as well.