# Sketch the function:?

Sketch the function:????

Hello everyone, I was asked the following question:
On the same axes, sketch the functions:
f(x) = 10+3x-x^2
g(x) = 3x^2-6x-5

Calculate the coordinates of the two points where the two lines cross. How far apart are the two points:

I don't even know how to start?
Am I supposed to use the quadratic formula and first sketch f(x) and then use the quadratic formula again to sketch g(x) on the same graph??

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You need to find where f(x) = g(x); those will be parts of the points where the two functions intersect. Set the functions equal to each other and solve for x. You should get two x values, one for each point. Plug them into either f(x) or g(x) and then you will get the y coordinates for the two points.

Lastly, do you know the distance formula for two points?

HallsofIvy
Homework Helper

What do you mean by " the quadratic formula and first sketch f(x) and then use the quadratic formula again to sketch g(x) on the same graph"? How does the quadratic formula help sketch a graph, except by finding the x-intrecepts. I recommend completing the square. After you $f(x)= a(x- b)^2+ c$ you can easily find the x- intercepts ($(x- b)^2= -c/a$ so $x= b\pm \sqrt{-c/a}$, the y-intercept ($ab^2+ c$) and, perhaps most importantly, the vertex $(b, c)$. With those points, it should be easy to sketch the graph.

f(x)=g(x)
10+3x-x^2=3x^2-6x-5
10+3x-x^2-3x^2=3x^2-3x^2-6x-5
10+3x-4x^2= -6x-5
10+3x-4x^2+6x=-6x+6x-5
-4x^2+3x+6x+10=-5
-4x^2+9x+10=-5
-4x^2+9x+10+5=-5+5
-4x^2+9x+15 =0

Applied the quadratic formula and got:
x =-1.1145
x=3.3645

Now,you told me to plug the two x values above to either f(x) or g(x), I don't know if you want me to do this:
f(x) =10+3x-x^2
f(x) =10+3(-1.1145)-(-1.1145)^2

or this:
f(x) =10+3x-x^2
f(x) =10+3(-1.1145)-(3.3645)^2

Help???????????????????

For the equation -4x2 + 9x + 15 = 0
use the quadratic formula to to get the exact answers for x with square roots and no rounding. Let's call them x1 and x2.

You chose f(x), so you want to first evaluate f(x1). That means plugging in only x1 in the function, as you did in the first one where you weren't sure, and not two different values as in the second one. The coordinates of that point of intersection will be (x1, f(x1)). Do the same thing with x2.