SBAC Practice Test Questions 31-32 (Day 176)

Today on her Mathematics Calendar 2019, Theoni Pappas writes:

If the volume of the cube is 8 cubic cm, then half the measure of angle BAC is _____ degrees.

[Here is the given info from the diagram: A, B, and C are vertices of the cube, chosen so that segments AB, AC, and BC are all face diagonals.]

Like yesterday, we will let h be the height of the cube -- that is, it's the length of any edge. Then we can let d be the length of a face diagonal.

Thus triangle ABC has three sides all of length d -- in other words, it's equilateral. All equilateral triangles are equiangular -- hence angle BAC is 60. Therefore half of this measure is 30 degrees -- and of course, today's date is the thirtieth.

Yes -- we could have easily determined that h = 2 and d = 2sqrt(2). I choose not to do so above in order to emphasize that the answer would have been 30 regardless of the size of the cube. Indeed, just like yesterday, the fact that the triangle is equilateral turns out to be key.

Question 31 of the SBAC Practice Exam is on right triangle trigonometry:

Consider this right triangle.

[The right angle is at S, ST = 21, and RT = 35.]

Determine whether each expression can be used to find the length of side RS. Select Yes or No for each expression.

                 Yes  No
35 sin(R)
21 tan(T)
35 cos(R)
21 tan(R)

Let's first look at the two involving 35 times something R. We notice that 35 is the hypotenuse and relative to angle R, the desired side RS is the adjacent side, so we need the cosine. Thus 35 cos(R) is yes, while 35 sin(R) is wrong.

Now we check out the 21 times tangent something. We see that the desired side RS is opposite angle T while 21 is adjacent to it. So the tangent of T is RS/21. Thus 21 tan(T) is yes, while 21 tan(R) is no.

Both the girl and the guy from the Pre-Calc class answer the first three parts correctly. But the guy leaves the last part blank while the girl answers it correctly. Since the guy knows that 21 tan(T) is yes, it's most likely an accidental omission on his part.

Question 32 of the SBAC Practice Exam is on the graphs of quadratic functions:

Given the function
y = 3x^2 - 12x + 9,

• Place a point on the coordinate grid to show each x-intercept of the function.
• Place a point on the coordinate grid to show the minimum value of the function.

To find the x-intercepts of this parabola, let's factor the function:

y = 3x^2 - 12x + 9

y = 3(x^2 - 4x + 3)

y = 3(x - 1)(x - 3)

So the x-intercepts are at (1, 0) and (3, 0). Since the function vanishes when x is 1 or 3, it follows that the axis of symmetry, hence the vertex, is at x = 2. There are other ways to find the vertex, but it's best just to take the mean of the x-intercepts if we've already found them.

Let's now find the y-value of the vertex:

y = 3x^2 - 12x + 9

y = 3(2)^2 - 12(2) + 9

y = 12 - 24 + 9

y = -3

Therefore the vertex is at (2, -3). The SBAC only requires students to plot these three points -- once again, I'm not sure whether the full parabola can be graphed on the computer interface.

The girl from the Pre-Calc class correctly graphs these three points and tries to graph a parabola, even though her graph looks more like a V-shape. But the guy, unfortunately, makes an error in factoring:

y = 3x^2 - 12x + 9

y = 3(x^2 - 4x + 3)

y = 3(x - 4)(x + 1)

and so his x-intercepts are at (4, 0) and (-1, 0). In other words, he graphs y = 3(x^2 -3x - 4) instead of the correct graph. This counts as both a sign error as well as confusing the needed sum and product during factoring.

So what does the guy do for his vertex? For the x-value, it appears that he wants to make his parabola look symmetrical. The mean of his two x-intercepts is 1.5. But the vertex he draws ends up being closer to x = 2, which is unwittingly the correct value. He seems to choose a random value of y -- his vertex is at (2, -4), one unit below the correct vertex of (2, -3).

Ironically, the guy's graph actually looks more like a parabola than the girl's graph. But the girl is the one who correctly find the intercepts and minimum.

SBAC Practice Exam Question 31

Common Core Standard:

CCSS.MATH.CONTENT.HSG.SRT.C.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems.

SBAC Practice Exam Question 32

Common Core Standard:

CCSS.MATH.CONTENT.HSF.IF.C.7.A
Graph linear and quadratic functions and show intercepts, maxima, and minima.

Commentary: The trig problem should be straightforward provided that the students know the definitions of the trig ratios. The parabola graphing will be tricky for the students who have trouble factoring the quadratic function.