#plz help!
47 messages · Page 1 of 1 (latest)
one box equals one unit
but wouldnt it be calculating the points (line to line) like usual
we know one root
and the y-intercept
i dont get how the zeros would be calculated
i am trying to first determine the function
Do you need the function? The question parts can be read off the graph by using the grid lines.
well the can you read the second root or the optimal point?
it's looks inaccurate thats why
if each box is 1x1, then it's just a matter of counting
okay
the second zero looks approximated -1.6 or so but this is not the most precise
would the optimal value be 13? or does that not make sense
how
idek because im counting the tiles all the way to the top
the optimal value would be y = 7 and something
ohh wait yes
and x = 1.5 approximated
okay thank u sm
wouldnt it be negative?
idk because the 5 is positive so would that make whatever is on top negatives?
i dont understand
This is the closest I can get with producing graph from the formula. adonhs is right, this thing's off a bit from the grid-lines. Counting requires guessing. Getting the formula took a system of 3 equations with 3 unknowns using 3 points read off the plot.
Worse, the axes don't math up with the grid. I wonder how precise the answers are expected to be for this problem.
And if I hit the 3 obvious points, I still miss the x-intercept on the left.
precisely my point thanks
A(1,7)
B(2,7)
C(5,0)
y = ax²+bx+c
a < 0
(1) 7 = a + b + c
(2) 7 = 4a + 2b + c
(3) 0 = 25a + 5b + c
Then we get a = -7/12, b = 7/4, c = 35/6
,w plot (-7/12)x^2 +(7/4)x + 35/6
with grid lines...
i mean since the coordinate system and parabola looks off it could be or not
anyway teachers like that should get suspended
can't even provide proper material resulting to the students wasting their time with confusion
the symmetry line's at x = 1.5. That's 3.5 units from x = 5. Go 3.5 units to left of 1.5 for other x-intercept and land on -2. The given parabola doesn't seem to match a true parabola 😦
wdym
If {5, 0} is one x-intercept, the other should be {-2, 0}. But the pic hits x-axis between -2 and -1
yea this is what I meant by the parabola being with the coordinate system off
like they were moved
that's a bad graph.
yea u r right
what's a student to do with a problem like this?
I'm tempted to just make my own parabola, solve that, and go given problem had errors. I can do this, here, I'll show you. And move on.
you can also go outside