Posts tagged with: graph

The midpoint formula tells you that the a segment with endpoints (a, b) and (c, d) will have a midpoint at ((c)/2,(d)/2)

So we know that (+ 9)/2 = 6 and (5 + y)/2 = 3. We can solve those!

(+ 9)/2 = 6
+ 9 = 12
x = 3

(5 + y)/2 = 3
5 + y = 6
y = 1

Therefore, x + y = 4.

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Draw this out. Start with the two points you’re given.

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Now remember that the shape is a rectangle, and that you’re told that point B is on the x-axis. The only way that happens is if B is at (5, 0). Point D, by the same logic, must be at (–3, 2).

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Now draw the rectangle and measure the lengths. The long ends have length 8, and the short ends have length 2.

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Therefore, the perimeter is 8 + 2 + 8 + 2 = 20.

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The thing to remember about functions is that they do the same thing to whatever is inside the parentheses. So don’t worry about the r vs. the h. They could use x, or a little star symbol, or whatever else they want. What matters is that the function f, as defined here, will equal zero when r = 4, or when r = –1.

When we’re told that f(h – 3) = 0, we can conclude that h – 3 must equal either 4 or –1. Therefore, h must equal either 7 or 2.

The graph below might help show what’s going on visually. When the thing in the parentheses next to the f (also known as the argument of the function) equals –1 or 4, the function equals zero. Therefore h – 3 must equal –1 or 4.

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A question from the May 2018 SAT (Section 4 #18)

kx + y = 1
y = -x² + k

In the system of equations above, k is a constant. When the equations are graphed in the xy-plane, the graphs intersect at exactly two points. Which of the following CANNOT be the value of k?

A. 3
B. 2
C. 1
D. 0

Because this comes in the calculator section, you might think that my recommendation is to graph them all. Indeed, if you go that route, you’ll get the answer, and it won’t take that long! However, if you notice that making k equal zero eliminates the x term in the first equation, you might not need to bother graphing anything else.

When k=0, the equations become y=1 and y=-x^2. The former is just a horizontal line above the x-axis, and the latter is a downward-facing parabola that intersects the origin. Those will never intersect, so zero cannot be the value of k.

Here are a couple questions from the old official SAT Subject Test Math I practice exam:

The function f is defined by f(x) is x^4 – 4x^2 + x + 1 for -5 ≤ x ≤ 5. In which of the following intervals does the minimum value of f occur?
A) -5 ≤ x ≤ -3
B) -3 ≤ x ≤ -1
C) -1 ≤ x ≤ 1
D) 1 ≤ x ≤ 3
E) 3 ≤ x ≤ 5

Can you solve w/o graphing?

Yes, but just before beginning I think it’s important to stress HOW USEFUL a graphing calculator is for the math Subject Tests. Anyone who’s prepping for those that doesn’t own one and can’t borrow one from school should find another way to get their hands on one they know how to use for test day.

To get this without graphing, you’re going to want to plug in values Just the integers should work. Note that x^4 will always be 0 or positive and -4x^2 will always be 0 or negative, but x will be both positive and negative, you should start by plugging in negative numbers.

(-5)^4-4(-5)^2-5+1=521

(-4)^4-4(-4)^2-4+1=189

(-3)^4-4(-3)^2-3+1=43

(-2)^4-4(-2)^2-2+1=-1

(-1)^4-4(-1)^2-1+1=-3

(0)^4-4(0)^2-0+1=-1

From there, you probably see that the minimum is around –1. So plug in a couple more values to see whether you want to choose B or C.

(-1.5)^4-4(-1.5)^2-1.5+1=-4.4375

(-0.5)^4-4(-0.5)^2-0.5+1=-0.4375

Yeah…gonna want to go with B.

Again, the graph is SO helpful here:

With that, you immediately see that the minimum is between –1 and –2.

Could you please explain SAT Past paper 3, section 4, question 12? I don’t really understand why the double root is considered a ‘distinct’ zero.

Thank you for your time!

Sure. Would it make you feel better if “distinct” were replaced with “unique”? Those terms are interchangeable here.

When the question says the function has “five distinct zeros,” that really just means there are five values that 1) don’t equal each other, and 2) make the function equal zero.

In other words, f(a) = 0, f(b) = 0, f(c) = 0, f(d) = 0, and f(e) = 0 for unique values abcd, and e.

In OTHER other words, the function f touches the x-axis five times. Sidestepping all of the above, all you really need to do is count the number of times each graph touches the x-axis.

Choices A and B touch the x-axis four times, and choice C touches the x-axis six times. Only choice D touches it five times.

Can you help me with the Official Test 4 Question 21?

Yep! I actually think it’s helpful to do a little drawing here. Below is my very rough (sorry, had to draw it on my laptop trackpad) drawing of the axes as described in the question, with year 2000 data on the x-axis and year 2010 data on the y-axis. I’ve also drawn the yx line.

Now let’s pick a set of bars and try to plot it. Looks like the bars for wood were at about 2.25 quadrillion Btu in 2000 and a little less than 2 quadrillion Btu in 2010.

Biofuels, on the other hand, were fairly small in 2000 (around 0.25) and larger in 2010 (say around 1.8).

Do you see what’s going on here? When consumption of an energy source grew from 2000 to 2010 (like biofuels), its dot will be above the yx line. When consumption declined from 2000 to 2010 (like wood), that dot will below the yx line.

Now you can stop plotting and just count the energy sources for which the 2010 bar is higher than the 2000 bar. I’m seeing three: biofuels, geothermal, and wind. So 3 is your answer.

Test 8 Section 4 #30

First, don’t be intimidated by all the visuals here. This one is not nearly as bad as it looks.

Start by identifying the maximum of the graph of f. Hopefully you agree with me that the greatest y-value the graph reaches is 3, which it reaches when x = 4. So we know that the maximum value of f is 3. Therefore, k = 3.

From there, all you need to do is read the table and find g(3). What does the table say in the g(x) column when it has a 3 in the x column? That’s right, g(3) = 6, so the answer is 6.

what is the difference between the maximum value of y=-x^4-2x^3+5 and the minimum value of y = x^4+x^2-4

This doesn’t seem like an SAT question—maybe Subject Test? Anyway, use your graphing calculator to find the maximum of the first function (it’s 107/16) and the minimum of the second function (it’s –4). Now subtract!

    \begin{align*}&\dfrac{107}{16}-\left(-4\right)\\=&\dfrac{107}{16}+\dfrac{64}{16}\\=&\dfrac{171}{16}\end{align*}

PSAT #2, Section 4, #29

I’d get this one by substituting. Since both equations are given to you in y = notation, set them equal to each other.

    \begin{align*}x-1&=x^2-4x+3\\0&=x^2-5x+4\\0&=(x-1)(x-4)\end{align*}

That tells you that x is either 1 or 4, so pick one and solve for what y must be in that case. Use the second equation because it’s simpler! For example, if x = 1, then the second equation tells you that y = 1 – 1 = 0. So in that case, the product of x and y is (1)(0) = 0.

The other possibility is that x = 4 and y = 3. In that case, the product of x and y would be (4)(3) = 12.

Note that although my first instinct here was to solve algebraically, this is also a good one to use your graphing calculator (if you have one). Just graph both equations, find either of the intersections, and multiply their coordinates to find the product of x and y.

Test 5, Section 4, Question 20. How do you get to the answer?

Get this by process of elimination.

A) Eliminate this because the question says the wheel is rolling at a constant rate (i.e., speed), so a graph of its speed would be a horizontal line.

B) Eliminate this because the question says the wheel is rolling at a constant rate along a straight, level path. Therefore, its distance from its starting point as a function of time would also be a straight line with a positive slope—the graph would get higher and higher the longer the wheel rolled.

C) Eliminate this because the mark on the rim will always be the same distance (the radius) from the center of the wheel.

D) By process of elimination this must be right. You might also recognize the similarities of the setup of this question to a model you might have seen in trigonometry when you were introduced to the sine curve:

via GIPHY

If r>0 and (9r/2)^(1/3) = (1/2) r, what is the value of r?

Two good ways to go here. First, the algebra:

    \begin{align*}\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}&=\dfrac{1}{2}r\\\\\left(\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}\right)^3&=\left(\dfrac{1}{2}r\right)^3\\\\\dfrac{9r}{2}&=\dfrac{1}{8}r^3\\\\36r&=r^3\end{align*}

One possible solution to the above is r = 0, but the question says that r > 0, so we can ignore that and move on:

    \begin{align*}36r&=r^3\\36&=r^2\\6&=r\end{align*}

We need not consider r = –6 because we know r is positive.

The other way to go is to graph both sides of the equation and find the intersection(s).

Because r is positive (my software made me use x in the graph above, as your calculator probably will, so just remember that xr), the only intersection that matters is (6, 3). At that intersection, x = 6, so we know that the value of r is 6.

Could you help me number 10 on page 70 please? I kept getting 0 as the answer

Yep! Here’s the question:

I think the easiest way to go here is to use the calculator to graph. Don’t worry about the ≥ signs, just graph the lines and remember that the ≥ means anything on or above the lines you see.

Now, remember that you’re looking for the lowest possible y-coordinate that’s on or above both of those lines. In other words, you’re looking for the y-coordinate of the intersection! Since you’ve already graphed the lines on your calculator, it’s probably easiest just to use the calculator to find the intersection: (–1.33, 2.33).

2.33 is the answer you’d grid in (or its fraction form: \dfrac{7}{3}).

Solving algebraically is also not too cumbersome. Since both equations are in y = form already, substitute and solve for x:

    \begin{align*}-4x-3&=\dfrac{1}{2}x+3\\\\-6&=\dfrac{9}{2}x\\\\\-\dfrac{4}{3}&=x\end{align*}

Once you have x, plug it into either equation to get y:

    \begin{align*}y&=-4\left(-\dfrac{4}{3}\right)-3\\\\y&=\dfrac{16}{3}-\dfrac{9}{3}\\\\y&=\dfrac{7}{3}\end{align*}

 

Hi Mike,

Can you tell me how to do #27 (Section 4, Practice Test 6). I know how to do it by completing the square but was wondering if there is an easier way.

thanks!
Rachel

If you have a calculator that can graph the given equation without rearranging it (like an Nspire) then the easiest way is just to do that.

       

If you don’t have one of those, though, then completing the square is 100% the way to go.

    \begin{align*}2x^2-6x+2y^2+2y&=45\\\\x^2-3x+y^2+y&=\dfrac{45}{2}\\\\x^2-3x+\dfrac{9}{4}+y^2+y+\dfrac{1}{4}&=\dfrac{45}{2}+\dfrac{9}{4}+\dfrac{1}{4}\\\\\left(x-\dfrac{3}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2&=25\end{align*}

Of course, since the formula for a circle is (x-h)^2+(y-k)^2=r^2, that 25 on the right side of the equation above tells you the radius is 5.

Practice Test 4, Section 3, Number 11 (No Calc)

I love this question because the fastest way to go involves almost no math. You just have to know a little bit about the shapes of lines and parabolas.

First, think about the parabola. Its equation is y=(2x-3)(x+9). From that we know it’s a parabola that opens up (the x^2 term will be positive) and has x-intercepts at \dfrac{3}{2} and -9. You should figure out its y-intercept, too, by plugging in zero for x:

    \begin{align*}y&=(2(0)-3)(0+9)\\y&=(-3)(9)\\y&=-27\end{align*}

Do a very rough drawing of that on your paper (forgive my MS Paint skillz, but your drawing can be sloppier than mine and still be plenty good enough):

Now do a rough drawing of the line. To do that, put it in slope-intercept form:

    \begin{align*}x&=2y+5\\x-5&=2y\\y&=\dfrac{1}{2}x-\dfrac{5}{2}\end{align*}

The important detail there is that the y-intercept is -\dfrac{5}{2}, which is above the parabola’s y-intercept of -27, so you know the line will intersect the parabola twice. Like so: