#Solving Equations with functions
1 messages · Page 1 of 1 (latest)
now draw the line y=12 (if u want to solve graphically)
done
do i need to do it on desmos
to do it in an exam u would need to solve it algebraically
ill rather do it that way then
so find the equation of each line segment, set it euqal to 12 and solve for each x
i understand i need to solve for x but where do i get the line segments from
cuz in the other questions it was easy cuz they gave it straight to you
on the diagram find the equation of the line between -10 and -4 just like at GCSE
and similarly for between -4 and 6
you know, like grandient = change in y / change in x etc..
so just find the equations of those 2 lines
for the equation at the bottom of the first screen shot why is it 5/2+12 and not 5/2-12?
because the equation of the line between -4 and 6 has its y-intercept at 12?
so h(x) = 12 and h(x) = 5/2x. we -12 to get h(x) on its own making it 5/2x-12 however it cant be -12 simply because the y-intercept is positive
you're confucing the difference between an "equation" and an "identity"
equation has = and identity has <> right?
what h(x) = 5x/2 + 12 is saying is that for values of x between -4 and 6 the function h(x) will ALWAYS take the value 5x/2 + 12.
what h(x) = 12 is saying is when is our function h(x) with its corresponding definition equal to 12
at a level they both just use "="
in reality when you define a function it should be h(x) := 5x/2 + 12
ok so there was no need for me to do any type of calculations cuz the 5/2x is simply a fact
basically ye
oh my i should of just looked down one more line they just equated it
however i dont underdand the top one cuz i did -4/6 = -2/3 then how did the y intercept become -6?
na fr