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you mean -16x^2?
and x-intercept means y=0, so -16x^2+22x+3=0
because 3 is prime we have (ax+3)(bx+1)=-16x^2+22x+3
how did you get that?
oh ok but why did you put 3 and 1 into the slots?
now, obvious cd=3, but 3 is prime which means the only possible answer is c=3,d=1, the other way round is the same, not a different answer
oh so when it says "What are the x-intercepts of the graph" they would be c=3,d=1?
we are not there yet, bear with me for a second
now expand LHS we have ab=-16,3b+a=22
we can recoginze that 8 and -2 is plausible answer
why cause they both go into 16?
since 3b+a is 22, a is -2, b is8
sorry what is your question?
Part A: What are the x-intercepts of the graph of the f(x)? Show your work.
now since y is 0, either -2x+3 is 0 or 8x+1 is 0
I dont understand "why cause they both go into 16?"
I said that because you said 8 and 2 are possible anwser
and 8 and 2 both go into 16
ab=-16, so you want one+ and one -
yea I get that part
8 and 2 is not possible since it would give the coeffecient before x^2 16,not -16
since 3b+a=22, we want to have 24-2, so a=-2 and b is 8
I forgot it was a negative ha
So now you should have factorized the quadratic equation
since y=0, either the first factor is 0 or the second, set them each to 0 and solve for x, thats your x intercept
oh ok so than what would be the anwser?
you can calculate it yourself, it is simple once you factored the quadratic eqaution
ok I got it now thanks and do you think you can help with part B and C?
Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points) Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points) @caozeyuan
maxmium because secondary derivative is negative
to find coord, you do first derivtive, set dy/dx=0, find, plug back in to find y
I am not sure if you finished calc I or not
well, a parabola is defined by 3 non collinear points, the two intercepts and the vertex are the three points, so yes it can be graphed
could you just help me with part C and ill be good
I am not sure about the standad steps, but I would draw a vertical line down to x axis from the vertex,this is the symmetry line of parabola, then connncet left intercept with vertex and right interest with vertex, make sure two curves are smooth and symmetrical
Well you did great thanks for helping me out :)