I do not understand question #4 pg 156 advanced systems of equations?
y=c(x^2) + d
In the systems of equations above, c and d are constants . For which of the following values of c and does the system of equations have no real solutions?
A) c=-6, d=6
B) c=-5, d=4
C) c=6, d=4
D) c=6, d=5
Please help, thank you.
Test 10 section 4 number 28
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?
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: 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?
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)
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?
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.
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 2 Section 4 #7
PSAT #1, Section 3, #13
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. 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!