Maths Functions Help: The function h:x → 2x^2+13x+15 xEℝ

A

asred9

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a) Find the range of h(x)
b Find an expression for h^-1 stating it domain & range.
c) Sketch h^-1 marking clearly and necessary values


h(x)=2x^2+13x+15
Turning Point: x=-b/2a=-13/2(2)=-3.25
sub x for -3.25
y=2(-3.25)^2+13(-3.25)+15
y=-6.125


Then I plotted the graph. with coordinates (-3.25,-6.25)


range h(x) = (h(x) is real, h(x) >=-6.125)


b) y=h(x)
y=2x^2+13x+15
after doing all necessary steps I ended with x=sqrt(y/2 +3.0625)-3.25


I found the domain which is range h^-1(x) = (x is real, x >=-6.125)
I couldnt find the range though


c)


The graph I cant plot.


Thanks hope I hear from you soon
 
asred9 said:
a) Find the range of h(x)
b Find an expression for h^-1 stating it domain & range.
c) Sketch h^-1 marking clearly and necessary values


h(x)=2x^2+13x+15
Turning Point: x=-b/2a=-13/2(2)=-3.25 Use grouping symbols. x = -b/(2a) = -13/[2(2)] = -3.25
sub x for -3.25
y=2(-3.25)^2+13(-3.25)+15
y=-6.125


Then I plotted the graph. with coordinates (-3.25,-6.25)


range h(x) = (h(x) is real, h(x) >=-6.125)


b) y=h(x)
y=2x^2+13x+15
after doing all necessary steps I ended with x=sqrt(y/2 +3.0625)-3.25


I found the domain which is range h^-1(x) = (x is real, x >=-6.125)
I couldnt find the range though

h(x) has no inverse function, because it's not a one-to-one function.
For instance, the graph of h(x) fails the horizontal line test.

c)


The graph I cant plot.
Click to expand...

\(\displaystyle h(x) = 2x^2 + 13x + 15 \ \ \) has no inverse.

But, \(\displaystyle \ \ h(x) = 2x^2 + 13x + 15,\ \ \ x \ge -3.25 \ \ \) does have an inverse.
 
Last edited:
asred9 said:
a) Find the range of h(x)
b Find an expression for h^-1 stating it domain & range.
c) Sketch h^-1 marking clearly and necessary values


h(x)=2x^2+13x+15
Turning Point: x=-b/2a=-13/2(2)=-3.25
sub x for -3.25
y=2(-3.25)^2+13(-3.25)+15
y=-6.125


Then I plotted the graph. with coordinates (-3.25,-6.25)


range h(x) = (h(x) is real, h(x) >=-6.125)


b) y=h(x)
y=2x^2+13x+15
after doing all necessary steps I ended with x=sqrt(y/2 +3.0625)-3.25


I found the domain which is range h^-1(x) = (x is real, x >=-6.125)
I couldnt find the range though


c)


The graph I cant plot.


Thanks hope I hear from you soon
Click to expand...
For part (b) you only have one part of the inverse function. That is
x = \(\displaystyle +\sqrt{\frac{y\, +\, 6.125}{2}}\, -\, 3.25\)
as one of the possible inverse functions or
x = \(\displaystyle -\sqrt{\frac{y\, +\, 6.125}{2}}\, -\, 3.25\)
as the other possible solution.

How small (and large) can y become [you have actually answered that in showing the domain]. When y is as small (as large) as it can be what is x?

For part (c), why can't you plot the graph? Just change your usual labels and make the horizontal axis y and vertical axis x. So, when you go to the left 6.125 units (y=-6.125), you go down 3.25 units (x=-3.25) for a point on the graph (actually the graph for either of the solutions).


EDIT: Note that you have two inverse functions here. In a strict sense, that would generally mean there is no inverse unless you restrict the domain of the original function. That restriction is reflected is the ranges of the two 'inverse functions'
 
Last edited:
lookagain said:
Do you have an e-mail address for the teacher so I may contact the teacher about
this problem?
Click to expand...

Unfortunately no. but i will sure tell her tomorrow.

Also I would be glad if you can help me on this other question since I have been trying hard since last night. thanks


Question 1:

A circle C is defined by the equation x^2+y^2-8x-4y+10=0
a) Find the coordinates of its center and its radius.
b) Show that the gradient m of the tangents of this circle which pass from the origin obey the equation 10(m^2+1)=(4m-2)^2
c) Hence show that these two tangents are perpendicular.

I managed to do the following:
x^2+y^2-8x-4y+10=0
(x^2-8x+16) + (y^2 -4y + 4) + 10 -16-4=0
(x-4)^2+ (y-2)^2= 10
The centre of the circle is at x=4, y=2

and the other one is linked:

https://answers.yahoo.com/question/index?qid=20150510060403AAfsf1d

I had like 15 questions and was given to us last Friday or Monday .... and those are the ones I got stuck in

Thanks again
 
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