lisztphisto
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the tangent to the curve y=x^2 + ax - 15 is horizontal at the point where x=4. find the value of a.
thx a million!
thx a million!
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derivative question (1 Viewer)
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watatank
=)
y=x^2 + ax - 15lisztphisto said:
gradient function: dy/dx = 2x + a
since tangent is horizontal, it must be a relative minimum at x=4.
to find relative min point 2x + a = 0
2(4) + a = 0
a = -8
lisztphisto
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cool! thanks!
ive got another question sorry..
A curve has equstion y = ax^3 + bx^2 +cx + d, a turning point at (0,5) , a point of inflexion when x=1/2 and crosses the x-axis at x=-1. Find the values of a,b,c and d.
i'd really appreciate it, if you could do this for me. thanks
ive got another question sorry..
A curve has equstion y = ax^3 + bx^2 +cx + d, a turning point at (0,5) , a point of inflexion when x=1/2 and crosses the x-axis at x=-1. Find the values of a,b,c and d.
i'd really appreciate it, if you could do this for me. thanks
-pari-
Active Member
sub (0,5) into the equation, you'll get d.
differentiate it, and equate y' to 0.
find the 2nd derivative, sub in (1/2) and equate this to 0.
you can sub in (0,-1) into the original equation as well..
you now have a series of equations, with some fiddling around, solve them simultaneously and you'll get your answers
differentiate it, and equate y' to 0.
find the 2nd derivative, sub in (1/2) and equate this to 0.
you can sub in (0,-1) into the original equation as well..
you now have a series of equations, with some fiddling around, solve them simultaneously and you'll get your answers
The information given is enough to derive 4 formulae which you solve simultaneouly for a, b, c and d.lisztphisto said:cool! thanks!
ive got another question sorry..
A curve has equstion y = ax^3 + bx^2 +cx + d, a turning point at (0,5) , a point of inflexion when x=1/2 and crosses the x-axis at x=-1. Find the values of a,b,c and d.
i'd really appreciate it, if you could do this for me. thanks
1) Since curve passes through (0,5), substitute x=0 and y=5 to find an equation involving a, b, c and d.
2) y'=3ax2+2bx+c
Now there is a turning at (0,5), so when x=0, y'=0. Substitute these values for an equation involving a, b and c.
3) y''=6ax+2b.
Given a point of inflexion at x=1/2, this means that y''=0 at this x-coordinate.
So y''=6a(1/2)+2b=0
3a+2b=0
I'm a little rusty with points of inflexion, so correct me if I'm wrong.
4) Curve crosses x-axis at x=-1 when y=0, so substitute into equation of curve to get another equation with a, b, c and d.