Verify that the slope of the tangent at this point is zero.

**GB_001** · October 26th 2008, 10:08 AM

From: Mr.DJ7 - Public Text WindowHello, I'm having trouble on this question, because my book didn't describe the solution properly:

For Each Function the point given is the maximum or minimum.
Use the difference quotient to verify that the slope of the tangent at this point is zero.

1. f(x)=0.5x^2 +6x + 7.5 Cords (-6, -10.5)

I also need help with this question:

Use an algebraic strategy to verify that the point given for each function is either a maximum or a minimum.

f(x)= x^2 - 4x + 5 Cords (2, 1)

**skeeter** · October 26th 2008, 10:48 AM

for the first problem ...

$\lim_{x \to -6} \frac{f(x) - f(-6)}{x - (-6)}$

$\lim_{x \to -6} \frac{\left(\frac{x^2}{2} + 6x + 7.5\right) + 10.5}{x +6}$

$\lim_{x \to -6} \frac{\frac{x^2}{2} + 6x + 18}{x + 6}$

$\frac{1}{2} \lim_{x \to -6} \frac{x^2 + 12x + 36}{x + 6}$

$\frac{1}{2} \lim_{x \to -6} \frac{(x+6)^2}{x + 6}$

$\frac{1}{2} \lim_{x \to -6} (x + 6) = 0$

for the second problem ...

$f(x)= x^2 - 4x + 5$

remember how to find the coordinates for the vertex of a parabola?

**GB_001** · October 26th 2008, 11:00 AM

Thankyou.
But I'm still kind of confused,
how did you get

$<br /> \frac{1}{2} \lim_{x \to -6} \frac{x^2 + 12x + 36}{x + 6}<br />$

And for the second one I Just substitute Y with f(x) and X with x?

**skeeter** · October 26th 2008, 02:50 PM

factor out 1/2 ...

$\frac{x^2}{2} + 6x + 18 = \frac{1}{2}(x^2 + 12x + 36)$

for the second problem, does $x = \frac{-b}{2a}$ ring a bell?

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