Question 963782
1.{{{dy/dx=10x-6}}}
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2. Slope of the tangent line is equal to the value of the derivative at (0,3).
{{{dy/dx=10x-6}}}
{{{m=dy/dx=10(0)-6}}}
{{{m=-6}}}
Use the point slope form of a line,
{{{y-3=-6(x-0)}}}
{{{highlight(y=-6x+3)}}}
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3. I'm assuming that {{{f(x)=5x^2-6x+3}}}
{{{f(h+3)=5(h+3)^2-6(h+3)+3}}}
{{{f(h+3)=5(h^2+6h+9)-6h-18+3}}}
{{{f(h+3)=5h^2+24h+30}}}
{{{f(3)=5(3)^2-6(3)+3}}}
{{{f(3)=45-18+3}}}
{{{f(3)=30}}}
So then,
{{{f(h+3)-f(3)=5h^2+24h+30-30}}}
{{{f(h+3)-f(3)=5h^2+24h}}}
{{{(f(h+3)-f(3))/h=5h+24}}}
{{{lim(h->0,(f(h+3)-f(3))/h=24)}}}
This is the definition of the derivative at {{{x=3}}}
As a check,
{{{dy/dx=10x-6}}}
{{{dy/dx=10(3)-6}}}
{{{dy/dx=30-6}}}
{{{dy/dx=24}}}