Answer:
Definition:
x -intercept : The point where the graph crosses the x-axis
Substitute in y=0 and solve for x
y-intercept : The point where the graph crosses the y-axis
Substitute in x=0 and solve for f(x) or y
1.
Given the function f(x) = [tex]\frac{1}{4}(x+2)(x-1)^2[/tex] .....[1]
to find x-intercept;
substitute y= 0 in equation [1];
[tex]\frac{1}{4}(x+2)(x-1)^2=0[/tex]
⇒ x+2 = 0 and [tex](x-1)^2 =0[/tex]
⇒ x =-2 and x= 1
Therefore, the x-intercept are; (-2,0) and (1 ,0)
Similarly, for y-intercept
Substitute x=0 in [1] to solve for y;
[tex]y=f(x)=\frac{1}{4} (0+2)(0-1)^2[/tex]
⇒[tex]y = \frac{1}{4}(2)(-1)^2[/tex]
Simplify;
y =[tex]\frac{1}{2}[/tex]
therefore, the y-intercept is, (0, [tex]\frac{1}{2}[/tex])
To find the relative extrema for the function f(x) = [tex]\frac{1}{4}(x+2)(x-1)^2[/tex];
Relative Extrema states that when the graph is turning around then there must be a horizontal tangent at that point, also we can say that the derivative value will be zero at that point, because a horizontal tangent has slope equal to 0.
As you can see in the Figure 1
Relative extrema of the function f(x) are (-1,1) and (1,0)
2.
Given the function h(x) =[tex]2x^3+5x^2-25x[/tex] .....[2]
to find x-intercept;
substitute y= 0 in equation [2];
[tex]2x^3+5x^2-25x=0[/tex] or
[tex]x(2x^2+5x-25)=0[/tex] or
[tex]x(x+5)(2x-5)=0[/tex]
Simplify:
x =0, x=-5 and [tex]x= \frac{5}{2}[/tex]
Therefore, the x-intercept are; (0,0), (-5,0) and ([tex]\frac{5}{2}[/tex],0)
Similarly, for y-intercept
Substitute x=0 in [2] to solve for y=h(x);
h(x) =[tex]2(0)^3+5(0)^2-25(0)[/tex]
h(x) =0
therefore, the y-intercept is, (0,0)
To find the relative extrema for the function h(x) =[tex]2x^3+5x^2-25x[/tex]
Relative Extrema states that when the graph is turning around then there must be a horizontal tangent at that point, also we can say that the derivative value will be zero at that point, because a horizontal tangent has slope equal to 0.
As you can see in the Figure 2
Relative extrema of the function h(x) are (-3.038, 66.019) and (1.371, -19.723)
