7) Find points P and Q on the parabola y =1-x², so that the triangle ABC formed by the x-axis and the tangent lines at P

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7) Find points P and Q on the parabola y =1-x², so that the triangle ABC formed by the x-axis and the tangent lines at P

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7 Find Points P And Q On The Parabola Y 1 X So That The Triangle Abc Formed By The X Axis And The Tangent Lines At P 1
7 Find Points P And Q On The Parabola Y 1 X So That The Triangle Abc Formed By The X Axis And The Tangent Lines At P 1 (39.27 KiB) Viewed 32 times
7 Find Points P And Q On The Parabola Y 1 X So That The Triangle Abc Formed By The X Axis And The Tangent Lines At P 2
7 Find Points P And Q On The Parabola Y 1 X So That The Triangle Abc Formed By The X Axis And The Tangent Lines At P 2 (33.11 KiB) Viewed 32 times
7) Find points P and Q on the parabola y =1-x², so that the triangle ABC formed by the x-axis and the tangent lines at P and Q is an equilateral triangle. B -0.5 -02 -0.4 0.5 Q 1 C HINT: Traingle ABC is an equilateral triangle so each angle of that triangle is 60 degrees. This makes slope of side AC to be -sqrt 3 and side BA to be sqrt3. Find derivative of y: 1-x^2 in order to find the slope of the parabola. Set the slope of the parabola equal to the two previously mentioned slopes in order to find the x-ordinate of Point P and Point Q. Next use this result to find the y-ordinates of Points P and Q.

Jaj 10% 10(a) sin(a) cos(a) at (3) f(a) S -y f(a,b)-z de 4. XX<a II ( x 120 d da S(X) Question 6 Without the aid of a graph, find the absolute maximum and minimum values of the function f(x)-36x+2665x+240x-675 +4534x-5836x¹-516x³ over the interval [0,4] HINT: Input your function into maple, find the first derivative and solve it for x after setting it to zero. Substitute the x-values obtained as well as your endpoints into the original functions to observe the y-values. Only subs x-values that lie within the closed interval given > f:x->36*x^6+2665*x^4+240 x-675+4534*x^2-5836*x*3-516*x^5; > df:ndiff(f(x),x); fx+36-x+2665-x+240-x-675+4534-²-5836-x-516- > £(0); > 2 (4); df 216x-2580x+10660-17508x² + 9068 x + 240 > cv: fsolve (df 0,x); > f(cv [2]); > f(cv [3]); > f(ev[4]); cv-0.02521970517, 1.030275579, 2.349186406, 3.800387934, 4.789814231 449,4473877 -452.25139 703.9551 -675 637 The absolute maximum value of the funtion is x-3.800387934; and the absolute minimum value of the funtion is x-0 (37) (38) (39) (40) (41) (42) (44)
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