Find the area of the triangle formed by the x- axis, the lines tangent and normal to the
graph of f(x) = 9 - x^2 at the point (2,5)
Step by Step please. 10 easy points.
graph of f(x) = 9 - x^2 at the point (2,5)
Step by Step please. 10 easy points.
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Graph thr data on graph paper.
f(x) = 9 - x^2 this is a parabola
Slope of a curve is the derivative of the function.
f'(x) = -2x at point ( 2,5) the slope is -2(2) = -4
Equation of the tangent line y-y1=m(x-x1) where (m = -4) and (y1 = 5)& ( x1 = 2)
thus y-5 = -4( x-2) multiply y-5 = -4x + 8
equation = y = -4x + 13 to find x intercept let y = 0: then 0 = -4x + 13 thus x = 13/4
x-axis point ( 13/4, 0)
The normal to the tangent line is perpendicular to the tangent line
perpendicular m2 = -(1/m1) wnere m1 = ( -4) thus m2 = - ( 1/-4) = 1/4
equation of normal line at point ( 2,5)
y-y1 = m( x-x1) then substitution of (2,5) gives y-5 = 1/4 (x-2)
y-5 = 1/4x - 2/4 equation is then y = 1/4x - 2/4 + 20/4 which is y= 1/4 x + 18/4
to find x intercept let y = 0 then 0= 1/4 x + 18/4 solve for x -18/4 = 1/4x
thus x= -18 and second point is (-18,0) Now we have three points on a triangle
Hero's formula states s= (a+b+c) / 2 the area of the triangle is A = sqrt [ s ( s-a)(s-b)(s-c)]
the distance formula will give the lengths of the sides
d= sqrt[( x2-x1)^2 + ( y2-y1)^2]
distance between ( -18,0) & (2,5) = 20.61
distance between (-18 ,0) & ( 13/4,0) = 21.25
distance between (2,5) & 13/4,0) = 5.15
thus s= (20.61+ 21.25+5.15)/2 = 23.5
a= 21.25 b= 5.15 c= 20.61
A= sqrt { s[ s-a)(s-b)(s-c)]} = sqrt[ 2770.422]= 52.63
A = 52.63
f(x) = 9 - x^2 this is a parabola
Slope of a curve is the derivative of the function.
f'(x) = -2x at point ( 2,5) the slope is -2(2) = -4
Equation of the tangent line y-y1=m(x-x1) where (m = -4) and (y1 = 5)& ( x1 = 2)
thus y-5 = -4( x-2) multiply y-5 = -4x + 8
equation = y = -4x + 13 to find x intercept let y = 0: then 0 = -4x + 13 thus x = 13/4
x-axis point ( 13/4, 0)
The normal to the tangent line is perpendicular to the tangent line
perpendicular m2 = -(1/m1) wnere m1 = ( -4) thus m2 = - ( 1/-4) = 1/4
equation of normal line at point ( 2,5)
y-y1 = m( x-x1) then substitution of (2,5) gives y-5 = 1/4 (x-2)
y-5 = 1/4x - 2/4 equation is then y = 1/4x - 2/4 + 20/4 which is y= 1/4 x + 18/4
to find x intercept let y = 0 then 0= 1/4 x + 18/4 solve for x -18/4 = 1/4x
thus x= -18 and second point is (-18,0) Now we have three points on a triangle
Hero's formula states s= (a+b+c) / 2 the area of the triangle is A = sqrt [ s ( s-a)(s-b)(s-c)]
the distance formula will give the lengths of the sides
d= sqrt[( x2-x1)^2 + ( y2-y1)^2]
distance between ( -18,0) & (2,5) = 20.61
distance between (-18 ,0) & ( 13/4,0) = 21.25
distance between (2,5) & 13/4,0) = 5.15
thus s= (20.61+ 21.25+5.15)/2 = 23.5
a= 21.25 b= 5.15 c= 20.61
A= sqrt { s[ s-a)(s-b)(s-c)]} = sqrt[ 2770.422]= 52.63
A = 52.63
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f'(2)=-4=slope of tngent at (2,5). Slope of normal is 1/4 (using mm'=-1). Eqn of tangent is y=-4x+13, which meets x-axis at x=13/4; Eqn of normal is y=(1/4)x+9/2, which meets x-axis at x=-18. So area of specified triangle is (1/2)(13/4 + 9/2)x5=155/8.