Please show all the works and details. I have attached my original works which i got only 2 out of 12 and because I am not able to explain well.

E 10 5 -5 0 5 10 -5 + N 2 X y = x N y = 4x - 4 Xamettmg Method The Tangent Line Without Calculus We want the slope of the tangent line to the graph of f(x) = x2 at the point (2, 4). We will need a point and the slope. We already have the point, so we need to nd the slope. For the sake of tradition, let's call the slope m. - Write the equation of a line through the point (2, 4) with the slope m in aging-slope erm. - Rearrange this equation 50 that y is isolated. 0 Write an equation whose solutions would be the intersection of f(x) and this line. 0 When does this equation have only one solution? Hint: Quadratic Formula. c Find this solution and write the equation of the tangent line. 0 Conrm your solution with graphing software like Desmos. Include an image with your submission. The Tangent Line Without Calculus One of the goals we have in this course is to use calculus to nd the tangent line to a curve at a point. Let's try to do one example without calculus. Let f (x) = x2. We hoven't completely dened the tangent line yet, but we hove seen that \"only touches the graph at one point' is not a good enough description. However. in this particular case. it is! Use this to convince yourself with geometric intuition that this is true. There is a slide or t e v a. If you move this the graph shows you the tangent line at the point (a, (:2). Imagine drawing any other line through this point. It can't be done without intersecting the graph at one other point. You can see this on the graph as well. There is a slider for the value it. This changes the slope of the line through the point (a. a2). When it = 1 it is the tangent line. For any other value the line intersects the parabola at two points (you may have to zoom waaaaay out to see this). There is one exception to this. but it will not concern us. Now we are going to nd the equation for the tangent line to this graph at the point (2. 4) algebraically. The process will also conrm our geometric intuition that there is only one linear function that intersects the graph only once at each point on the graph. You may know how to nd this line with calculus if you have seen derivatives before. Do not use calculus to do this. Follow the method I outline below We have plenty of time for calculus during the rest of the quarter! '5 [ :J.:i.ii.n!i}' .ID- M. * F (x ) = X2 $ F ( x) = x2 @ point (2 14) ( yz -y1) = m (x2- x1) ( y- 4 ) = (4) ( x-2) How did you determine this ? ( y- 4 ) = 4x - 8 .: m= 4 The entire point y - 4 = 4 x - 8 b = - 4 of The project was 9 = 4x - 8+4 determing this and y = 4 x - 4 you skipped that .. y= y 4 x - 4 = x 2 * 2 - 4 x +4 = 0 Li(ar" +bxtc ) - 9= 1 6= - 4 ( = 4 = 0 .: 62 - 4ac = 0 ( because its tangent ) (- 4 ) 2 - 4 ( 1) ( 4 ) = 0 14 - 14 = 0 0 = 0 .: 9= 4 x -4 is the equation OF the tangent line