Answer to Question #260348 in Calculus for jall

Question #260348

Sketch a graph that has:

  1. 1 turning point, a maximum value and a y-intercept of (0,5). (3 pts)
  2. End behaviours of III → I, y-intercept of 2 and 2 x-intercepts. (3 pts)
  3. End behaviour of II → I, 1 x-intercept.
1
Expert's answer
2021-11-03T15:56:51-0400

1. A graph of parabola has 1 turning point. Let f(x)=ax2+bx+c.f(x)=ax^2+bx+c.

If a function has a maximum value, then a<0.a<0.

y-intercept: f(0)=a(0)2+b(0)+c=cf(0)=a(0)^2+b(0)+c=c

Example;


f(x)=x2+2x+5f(x)=-x^2+2x+5



2. As x,f(x)x\to-\infin, f(x)\to-\infin

As x,f(x)x\to\infin, f(x)\to\infin

If x1,x2x_1, x_2 are x-intercepts then


f(x)=a(xx1)(xx2)2,a>0f(x)=a(x-x_1)(x-x_2)^2, a>0

f(0)=ax1(x2)2=2f(0)=-ax_1(-x_2)^2=2

Let a=2,x1=1,x2=1a=2, x_1=-1, x_2=1


f(x)=2(x+1)(x1)2f(x)=2(x+1)(x-1)^2



3. As x,f(x)x\to-\infin, f(x)\to\infin

As x,f(x)x\to\infin, f(x)\to\infin

If x1x_1 is x-intercepts then


f(x)=a(xx1)2,a>0f(x)=a(x-x_1)^2, a>0

Let a=1,x=1a=1, x=1


f(x)=(x1)2f(x)=(x-1)^2




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