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anonymous

  • 5 years ago

Let f(x)=7x^2-4x+6. Then the quotient (f(5+h)-f(5))/h can be simplified to ah+b for a=? and b=?

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  1. amistre64
    • 5 years ago
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    are we looking for the equation of the tangent line at f(5)?

  2. anonymous
    • 5 years ago
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    I'm not sure, those are the only instructions I was given and I was never shown any sample problems, but I think that would make sense

  3. amistre64
    • 5 years ago
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    have you learned to do derivatives yet? or is that what you getting up to?

  4. anonymous
    • 5 years ago
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    ive learned

  5. amistre64
    • 5 years ago
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    good, the slope of the line is the derivative of the function at f'(5).

  6. amistre64
    • 5 years ago
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    as far as I know, "a" would equal f'(5)

  7. anonymous
    • 5 years ago
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    I dont think he is working on a derivative otherwise he would be taking the limit as h approaches 0

  8. amistre64
    • 5 years ago
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    b might = f(5) - f'(5)(h)... nadeems got a valid point :)

  9. anonymous
    • 5 years ago
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    i understand how to find the derivative, i'm just confused what the ah+b is representing in all of this

  10. anonymous
    • 5 years ago
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    f(5 + h) - f(5) = 7(5 + h)^2 - 4(5 + h) + 6 - 7(25) + 4(5) - 6 = 7(25 + 10h + h^2) - 20 - 4h - 155 = 175 + 70h +7h^2 -175 -4h 7h^2 + 66h [f(5 + h) - f(5)] = (7h^2 + 66h)/h = 7h +66

  11. amistre64
    • 5 years ago
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    might be a dummy form of the line equation... maybe h needs to be kept in as the variable?

  12. anonymous
    • 5 years ago
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    \[f(x)=7x^2-4x+6\] \[f(5+h)=7(5+h)^2-4(5+h)+6\] \[f(5)=7(5^2)-4(5)+6\]

  13. anonymous
    • 5 years ago
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    now plug in these values in to the function: \[\frac{f(5+h)-f(5)}{h}\rightarrow \frac{7(h^2+10h+25)-20-4h+6-159}{h}\]

  14. anonymous
    • 5 years ago
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    i see now

  15. anonymous
    • 5 years ago
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    now simplify: \[\frac{7h^2+66h}{h}\rightarrow \frac{7h^2}{h}+\frac{66h}{h}\rightarrow 7h+66\]

  16. anonymous
    • 5 years ago
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    so a=7 and b=66

  17. anonymous
    • 5 years ago
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    great, thanks!

  18. anonymous
    • 5 years ago
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    no problem

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