As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of \(\displaystyle f(x)=5x^{2}+9x-12\) and substitute in the \(\displaystyle x\)-value of the point \(\displaystyle (-1,3)\). 

Using the Power Rule \(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\), \(\displaystyle f'(x)=(2)5x^{2-1}+(1)9x^{1-1}-(0)12=10x+9\), Subbing in \(\displaystyle x=-1\), we get \(\displaystyle f'(-1)=10(-1)+9=-10+9=-1\).

As slope is defined as the derivative of a given function at a given point, we will need to take the derivative of \(\displaystyle f(x)=x^{2}-7x-6\) and substitute in the \(\displaystyle x\)-value of the point \(\displaystyle (2,9)\). 

Using the Power Rule \(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\), 

\(\displaystyle f'(x)=(2)x^{2-1}-(1)7x^{1-1}-(0)6=2x-7\).

Swapping in \(\displaystyle x=2\), we get \(\displaystyle f'(2)=2(2)-7=4-7=-3\).

Given the function \(\displaystyle f(x)=7x^{2}-2x+9\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to find its derivative:

\(\displaystyle f'(x)=(2)7x^{2-1}-(1)2x^{1-1}+(0)9=14x-2\).

Plugging in the \(\displaystyle x\)-value of the point \(\displaystyle (2,5)\) into \(\displaystyle f'(x)\), we get 

\(\displaystyle f'(2)=14(2)-2=28-2=26\).

Given the function \(\displaystyle f(x)=5x^{2}+6x-12\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to find its derivative:

\(\displaystyle f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6\).

Plugging in the \(\displaystyle x\)-value of the point \(\displaystyle (-1,-1)\) into \(\displaystyle f'(x)\), we get 

\(\displaystyle f'(-1)=10(-1)+6=-10+6=-4\).

Given the function \(\displaystyle f(x)=10x^{2}-9x+1\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to find its derivative:

\(\displaystyle f'(x)=(2)10x^{2-1}-(1)9x^{1-1}+(0)=20x-9\).

Plugging in the \(\displaystyle x\)-value of the point \(\displaystyle (-2,7)\) into \(\displaystyle f'(x)\), we get 

\(\displaystyle f'(-2)=20(-2)-9=-40-9=-49\).

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

\(\displaystyle y'=(2x+4)(3)+(3x+2)(2) = 6x+12 + 6x+4= 12x+16\)

Substitute \(\displaystyle x=1\).

\(\displaystyle y'(1)= 12(1)+16=28\)

The derivative of \(\displaystyle \small f(x)=2(x^2-1)^6\) can be computed using the chain rule:

\(\displaystyle \small \small f'(x)=2\cdot 6\cdot 2x(x^2-1)^5=24x(x^2-1)^5\)

so now we just plug in \(\displaystyle \small x=2\):

\(\displaystyle \small \small \small f'(2)=24\cdot2(2^2-1)^5=48\cdot 3^5=11,664\)

Slope is defined as the derivative of a function at a given point. By the Power Rule, 

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\),  

\(\displaystyle f'(x)=(2)x^{2-1}-(1)7x^{1-1}+(0)6=2x-7\).

At \(\displaystyle (2,3)\) the \(\displaystyle x\)-value is \(\displaystyle x=2\), so the slope 

\(\displaystyle f'(2)=2(2)-7=4-7=-3\).

Slope is defined as the derivative of a function at a given point.

By the Power Rule, 

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\),  

\(\displaystyle f'(x)=(2)5x^{2-1}-(1)7x^{1-1}+(0)2=10x-7\).

At \(\displaystyle (-3,-1)\) the \(\displaystyle x\)-value is \(\displaystyle x=-3\), so the slope 

\(\displaystyle f'(-3)=10(-3)-7=(-30)-7=-37\).

Slope is defined as the derivative of a function at a given point.

By the Power Rule, 

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\),  

\(\displaystyle f'(x)=(2)2x^{2-1}-(1)9x^{1-1}+0(5)=4x-9\).

At \(\displaystyle (4,-1)\) the \(\displaystyle x\)-value is \(\displaystyle x=4\), so the slope 

\(\displaystyle f'(4)=4(4)-9=16-9=7\).