In the xy-plane,the graph of the function h is a line.If h(-1)=4 and h(5)=1,what is the value of h(0) ?
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OpenStudy (anonymous):
I think it will b helpful if you consider h as f(x)
that is H(-1)=4
means... when value of x is -1 ; y value (f(x)= is 4...
and soon.. construct the the line equation using two point form.. then find value for h(0)
OpenStudy (hba):
(-1,4) and (5,1)
OpenStudy (anonymous):
We can find 2 points from h(-1)=4 and h(5)=1
OpenStudy (anonymous):
yes.. you got it
OpenStudy (anonymous):
now construct line equation
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OpenStudy (hba):
What next ?
OpenStudy (anonymous):
Use the points to find the slope
OpenStudy (hba):
y-y1=m(x-x1) ??
OpenStudy (anonymous):
yes
OpenStudy (hba):
m=y2-y1/x2-x1
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OpenStudy (anonymous):
yes
OpenStudy (hba):
m=-1/2
OpenStudy (hba):
y+1=-1/2(x-4)
OpenStudy (anonymous):
??? what is x1 and y1
OpenStudy (hba):
(x1,y1)=(-1,4)
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OpenStudy (anonymous):
y+1=-1/2(x-4) is it y-y1=m(x-x1) ??? can you please check...
OpenStudy (anonymous):
your slope is correct...but the point you took is wrong
OpenStudy (hba):
No it is correct :)
OpenStudy (hba):
Do i have to take the other two points ?
OpenStudy (anonymous):
is it...I have doubt..
y1 is 4 and x1 is-1... but in the abopve equation you took y1 as -1 and x1 as 4
amI right?
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OpenStudy (hba):
My bad :(
OpenStudy (anonymous):
:) correct it
OpenStudy (hba):
y-4=-1/2(x+1)
OpenStudy (anonymous):
yes.. now find h(0)
OpenStudy (hba):
Ok so x=0
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