5. (5pts) find the slope ( y^{prime}(x) ) of the tangent line to the curve of ( y(x)=a x^{2}+b x+c ) where ( a, b ), and ( c ) are constants. recall that the slope of the tangent line is given by ( y^{prime}(x)=\frac{y(x+delta x)-y(x)}{delta x} ) then let ( delta x
ightarrow 0 ) show all of your work. 6. (5pts) similarly, the velocity ( v(t) ) is the instantaneous rate of change (slope of the tangent line) of ( x(t) ) ( v(t)=\frac{x(t+delta t)-x(t)}{delta t} ) then let ( delta t
ightarrow 0 ) find the velocity ( v(t) ) given that ( x(t)=3 t^{2}+4 t ).

Question

Accepted Answer

### Problem 5
# Explanation:
## Step1: Substitute $y(x)$ into the formula
Given $y(x)=Ax^{2}+Bx + C$, then $y(x+\Delta x)=A(x + \Delta x)^{2}+B(x+\Delta x)+C$.
Expand $y(x+\Delta x)$:
$$
\begin{align*}
y(x+\Delta x)&=A(x^{2}+2x\Delta x+\Delta x^{2})+Bx + B\Delta x+C\
&=Ax^{2}+2Ax\Delta x+A\Delta x^{2}+Bx + B\Delta x+C
\end{align*}
$$
Now, $y(x+\Delta x)-y(x)=(Ax^{2}+2Ax\Delta x+A\Delta x^{2}+Bx + B\Delta x+C)-(Ax^{2}+Bx + C)$
$$
\begin{align*}
y(x+\Delta x)-y(x)&=2Ax\Delta x+A\Delta x^{2}+B\Delta x\
\end{align*}
$$
## Step2: Calculate $\frac{y(x+\Delta x)-y(x)}{\Delta x}$
$$
\frac{y(x+\Delta x)-y(x)}{\Delta x}=\frac{2Ax\Delta x+A\Delta x^{2}+B\Delta x}{\Delta x}=2Ax + A\Delta x+B
$$
## Step3: Take the limit as $\Delta x
ightarrow0$
$$
y^{\prime}(x)=\lim_{\Delta x
ightarrow0}\frac{y(x+\Delta x)-y(x)}{\Delta x}=\lim_{\Delta x
ightarrow0}(2Ax + A\Delta x+B)
$$
Since $\lim_{\Delta x
ightarrow0}(A\Delta x) = 0$, we have $y^{\prime}(x)=2Ax + B$

# Answer:
$y^{\prime}(x)=2Ax + B$

### Problem 6
# Explanation:
## Step1: Substitute $x(t)$ into the formula
Given $x(t)=3t^{2}+4t$, then $x(t+\Delta t)=3(t + \Delta t)^{2}+4(t+\Delta t)$
Expand $x(t+\Delta t)$:
$$
\begin{align*}
x(t+\Delta t)&=3(t^{2}+2t\Delta t+\Delta t^{2})+4t + 4\Delta t\
&=3t^{2}+6t\Delta t+3\Delta t^{2}+4t + 4\Delta t
\end{align*}
$$
Now, $x(t+\Delta t)-x(t)=(3t^{2}+6t\Delta t+3\Delta t^{2}+4t + 4\Delta t)-(3t^{2}+4t)$
$$
\begin{align*}
x(t+\Delta t)-x(t)&=6t\Delta t+3\Delta t^{2}+4\Delta t
\end{align*}
$$
## Step2: Calculate $\frac{x(t+\Delta t)-x(t)}{\Delta t}$
$$
\frac{x(t+\Delta t)-x(t)}{\Delta t}=\frac{6t\Delta t+3\Delta t^{2}+4\Delta t}{\Delta t}=6t + 3\Delta t+4
$$
## Step3: Take the limit as $\Delta t
ightarrow0$
$$
v(t)=\lim_{\Delta t
ightarrow0}\frac{x(t+\Delta t)-x(t)}{\Delta t}=\lim_{\Delta t
ightarrow0}(6t + 3\Delta t+4)
$$
Since $\lim_{\Delta t
ightarrow0}(3\Delta t) = 0$, we have $v(t)=6t + 4$

# Answer:
$v(t)=6t + 4$

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QUESTION IMAGE

Question

Explanation:

Problem 5

Step1: Substitute \(y(x)\) into the formula

Step2: Calculate \(\frac{y(x+\Delta x)-y(x)}{\Delta x}\)

Step3: Take the limit as \(\Delta x

Step1: Substitute \(x(t)\) into the formula

Step2: Calculate \(\frac{x(t+\Delta t)-x(t)}{\Delta t}\)

Step3: Take the limit as \(\Delta t

Answer:

Problem 6