Mixing
Let $f(x)= 2x^3 +Ax^2 +4x -5$; find $A$ given $f(2) =5$
$begingroup$
Let
$$f(x)= 2x^3 +Ax^2 +4x -5$$
Find $A$ given $f(2) =5$.
If I could have someone show me how to solve this it would be great
algebra-precalculus
edited Jan 31 at 0:01
Surb
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
$endgroup$
add a comment |
$begingroup$
Let
$$f(x)= 2x^3 +Ax^2 +4x -5$$
Find $A$ given $f(2) =5$.
If I could have someone show me how to solve this it would be great
algebra-precalculus
edited Jan 31 at 0:01
Surb
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
$endgroup$
add a comment |
$begingroup$
Let
$$f(x)= 2x^3 +Ax^2 +4x -5$$
Find $A$ given $f(2) =5$.
If I could have someone show me how to solve this it would be great
algebra-precalculus
edited Jan 31 at 0:01
Surb
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
$endgroup$
Let
$$f(x)= 2x^3 +Ax^2 +4x -5$$
Find $A$ given $f(2) =5$.
If I could have someone show me how to solve this it would be great
algebra-precalculus
algebra-precalculus
edited Jan 31 at 0:01
Surb
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
edited Jan 31 at 0:01
Surb
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
edited Jan 31 at 0:01
Surb
38.4k94478
edited Jan 31 at 0:01
Surb
38.4k94478
edited Jan 31 at 0:01
Surb
38.4k94478
38.4k94478
asked Jan 30 at 23:55
RicardoRicardo
1
asked Jan 30 at 23:55
RicardoRicardo
1
asked Jan 30 at 23:55
RicardoRicardo
1
1
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$implies 16 + 4A + 8 - 5 = 5$$ $$A = cdots?$$
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
$endgroup$
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$implies 16 + 4A + 8 - 5 = 5$$ $$A = cdots?$$
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
$endgroup$
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
add a comment |
$begingroup$
Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$implies 16 + 4A + 8 - 5 = 5$$ $$A = cdots?$$
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
$endgroup$
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
add a comment |
$begingroup$
Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$implies 16 + 4A + 8 - 5 = 5$$ $$A = cdots?$$
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
$endgroup$
Hint: The solution is very straightforward. Since we know that $f(2) = 5$, then obviously, $$f(2) = 2(2)^3 + A(2)^2 + 4(2) - 5 = 5$$ $$implies 16 + 4A + 8 - 5 = 5$$ $$A = cdots?$$
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
answered Jan 30 at 23:59
Decaf-MathDecaf-Math
3,422926
3,422926
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
add a comment |
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
1
1
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
$begingroup$
Ohhhh so A=-7/2
$endgroup$
– Ricardo
Jan 31 at 0:09
add a comment |
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