Questions: Use the definition (mtan =lim h rightarrow 0 fracf(a+h)-f(a)h) to find the slope of the line tangent to the graph of (f) at (P). Determine an equation of the tangent line at (P). (f(x)=7+4 x^2 ; P(0,7)) (mtan =) (square) (Type an integer or a fraction.)

Use the definition (mtan =lim h rightarrow 0 fracf(a+h)-f(a)h) to find the slope of the line tangent to the graph of (f) at (P).
Determine an equation of the tangent line at (P).
(f(x)=7+4 x^2 ; P(0,7))
(mtan =) (square) (Type an integer or a fraction.)
Transcript text: Use the definition $m_{\tan }=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$ to find the slope of the line tangent to the graph of $f$ at $P$. Determine an equation of the tangent line at $P$. \[ f(x)=7+4 x^{2} ; P(0,7) \] $m_{\tan }=$ $\square$ (Type an integer or a fraction.)
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Solution

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Find the slope of the line tangent to the graph of f f at point P P .

Calculate f(a+h) f(a+h) and f(a) f(a) for a=0 a = 0 .

We have f(a+h)=f(0+h)=7+4(h)2=4h2+7 f(a+h) = f(0+h) = 7 + 4(h)^2 = 4h^2 + 7 and f(a)=f(0)=7 f(a) = f(0) = 7 .

Use the limit definition to find the slope mtan m_{\tan} .

Using the limit definition, we find:
mtan=limh0f(a+h)f(a)h=limh0(4h2+7)7h=limh04h2h=limh04h=0. m_{\tan} = \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \rightarrow 0} \frac{(4h^2 + 7) - 7}{h} = \lim_{h \rightarrow 0} \frac{4h^2}{h} = \lim_{h \rightarrow 0} 4h = 0.

The slope of the tangent line is 0 \boxed{0} .

Determine an equation of the tangent line at point P P .

Use the point-slope form of the line equation.

The point-slope form is given by yy1=m(xx1) y - y_1 = m(x - x_1) . Here, m=0 m = 0 and (x1,y1)=(0,7) (x_1, y_1) = (0, 7) . Thus, the equation becomes:
y7=0(x0)    y=7. y - 7 = 0(x - 0) \implies y = 7.

The equation of the tangent line is y=7 \boxed{y = 7} .

The slope of the tangent line is 0 \boxed{0} .
The equation of the tangent line is y=7 \boxed{y = 7} .

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