<h3>
Answer: y = x+1</h3>
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Explanation:
f(x) = x^3 - 2x + 3
f ' (x) = 3x^2 - 2 ..... apply the power rule
f ' (1) = 3(1)^2 - 2 ... plug in x coordinate of given point
f ' (1) = 1
If x = 1 is plugged into the derivative function, then we get the output 1. This means the slope of the tangent line at (1,2) is m = 1. It's just a coincidence that the x input value is the same as the slope m value.
Now apply point slope form to find the equation of the tangent line
y - y1 = m(x - x1)
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x - 1 + 2
y = x + 1 is the equation of the tangent line.
The graph is shown below. I used GeoGebra to make the graph.
X1-x2/y1-y2 is what you need for ordered pair
Answer:
I believe the answer is 9 years old or 8 years old.
Answer:
If the two beside the p in d is an exponent I believe that would be correct. if not sorry
![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.