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find a so that the point (-1, 2) is on the graph of f(x)=ax^2+5
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since (-1,2) is on the graph f(x)=ax^2+5 you can put f(x)=2, and x=-1 what u get ?
2=a(-1)+5
2=a(-1)^2+5 find a from here
don't miss ^2
a = ..
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... ?
3
?
very close,but not correct 2=a+5 a=?
-3
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that is correct value of a
thank you again i also have a question just like the one we did a question ago. do you have time?
few minutes only...
4x^2 + 25y^2 = 100
intercepts?
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yes there will be four i know.
25,0 -25,0 for x?
nopes
y 0 -4 0 4
nopes
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darn.. can u help?
u got x^2 =25 right ?
yes
5
-5
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those will be x ints? (-5,0) (5,0)
\(\checkmark\)
0,-2 0,2
y
that is also \(\checkmark\)
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thank u so much. when r u on next? u should be my reg helper.
idk, i come online when i m free...
oh ok.
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