# Is there a way to solve this system of equations without using the quadratic formula (or graphing)?

Is there a way to solve this system of equations without using the quadratic formula (or graphing)?

f(x) = –2/3 x + 4
g(x) = 3(x + 2)^2 – 4

How many solutions does the system above have?

# What is the x-coordinate of the vertex of the graph of y = -6x^2 + 3x + 8

Huge shortcut here if you just know that for a parabola in standard ax^2 + bx + c form, the x-coordinate of the vertex will be at –b/(2a). In this case, that means it’s at –3/(2(–6)) = 3/12 = ¼. from Tumblr https://ift.tt/2CRW93a

# PWN the SAT Parabolas drill explanation p. 325 #10

PWN the SAT Parabolas drill explanation p. 325 #10: The final way to solve: If we are seeking x=y, since the point is (a,a), why can you set f(x) = 0? You start out with the original equation in vertex form, making y=a and x=a, but halfway through you change to y=0 (while x is still = a). How can we be solving the equation when we no longer have a for both x and y?

# A tennis ball is thrown upward from the ground and its height, h, is given by the equation h=22t – t^2. Some kids are sitting on the roof of a building that stands 21 feet tall. If the kids are sitting in such a position that they cannot see the ball until it reaches the height of the roof, for how many seconds of the tennis ball’s flight can the kids see the ball? A- 18 B-20 C-21 D-22

Basically, the question is: how many seconds is h greater than 21? (This tennis ball is being thrown on a planet other than Earth, by the way. I challenge anyone to throw a tennis ball that stays in the air anywhere near as long as this one does.) To figure it out, solve for the (more…)

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

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

# Can you please explain Question 11, Test 6, section 3?

Hi Mike, Can you please explain Question 11, Test 6, section 3 ? I know the parabola opens downward, but I’m confused after that. Thanks.

# PWN p. 157 #8

In PWN p. 159 (p. 157 later printing) #8
In the xy-plane, where a and b are constants, the graphs …

The question does not specify that a and b are positive values. If one or both were negative, wouldn’t that change the answer?

# Test 6 Section 3 #13

Test 6 Section 3 #13

# Test 2 Section 4 #7

Test 2 Section 4 #7

# Question number 6 from the Parabola chapter

Hi, Mike! Can you explain the second way we can approach question number 6 from the Parabola chapter? (The two points where the higher y-coordinate is also farther from the line of symmetry.) It would be great if you can provide an example. Thank you.

# Parabola D in the xy-plane has equation x – 2y^2 – 8y – 11 = 0…

Parabola D in the xy-plane has equation x – 2y^2 – 8y – 11 = 0. Which equation shows the x-intercept(s) of the parabola as constants or coefficients?

A) x = 2y^2 + 8y + 11
B) x = 2(y + 2)^2 + 3
C) x – 3 = 2(y + 2)^2
D) y = – √(x – 3)/2 – 2

The answer is A) which makes me think this is really just a “do you know the definition of this term” kind of question (similar to official practice test 4.4.28). Can you explain for those weak in this? TIA!

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

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

# How do you do Test 5 Section 4 #35?

How do you do Test 5 Section 4 #35?

# Test 6 calculator section #34 please!

Test 6 calculator section #34 please!