Answer to Question #260348 in Calculus for jall

Question

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.

Answer 1

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

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

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

Example;

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

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

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

If $x_1, x_2$ are x-intercepts then

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

$$f(0)=-ax_1(-x_2)^2=2$$

Let $a=2, x_1=-1, x_2=1$

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

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

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

If $x_1$ is x-intercepts then

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

Let $a=1, x=1$

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