Can anyone graph these two linear functions?

Question

Q1. f(x) = –3x + 7 and
Q2. f(x) = –3x2 + x – 5

I haven't had to do anything like this in years and it's driving me insane. Any help would be appreciated, I know it's impossible to download a graph to "yahoo answers", step by step instructions would be wonderful. Please explain as simply as possible "Algebra for Dummies" style .... thank you in advance, I look forward to reading some interesting responses.

Whoopsi daisy, the second question is actually f(x) = -3x (to the power of two) + x -5, sorry about that

Answer 1

The simplest way is connect dots when you input x to find y or f(x), and input y or f(x) to find x.

First fuction, let call f(x) = y, therefore,
y = -3 x + 7, from looking at this, we know it's a straight line, and please don't ask how I know. Also, a straight line with a slope other than 0, which in this case, it is -3, the line will intercept the x-axis and y-axis some time. Enough said, let's x= 0, and find y
x=0; y = -3 * 0 + 7 ; y = 7. From there, we have a point on the graph (0,7). This is called the y-axis intercept. Let's find x-axis intercept. For x-axis intercept, y must equal to 0, so let's y = 0
y = 0; 0 = -3x + 7, and isolate x on 1 side,
7 = -3x ; => -7/3 = x. So the point on the graph is (-7/3,0). It's a straight line. So, you can connect the two points on the graph (0,7) & (-7/3,0) w/ a straight line and you got yourself a graph.

Q2. For this problem, use the same method, only that you know this is a up-side-down parabolla. So, find f(x) or y by simply substitute in x = 0, 1, -1, 2, -2, ... Then connect the dots on the graph. For instance,

let x = 0, y = -3*0^2 + 0 -5 = -5 => (0,-5)
let x = 1, y = -3*1^2 + 1 - 5 = -7 => (1,-7)
let x = -1, y = -3*(-1)^2 -1 - 5 = -9 => (-1,-9)

And so forth. I hope this will help you. This is the simpliest way I can think of without having to bust out formula and the whole 9-yard

Answer 2

Q1. f(x) = –3x + 7 and
Q2. f(x) = –3x2 + x – 5
Let y = f(x) so we get:
y = –3x + 7 <-- Eq 1
y = –3x2 + x – 5 <-- Eq2
Now you are ready to plot on x-y coordinate system
In Eq 1 let x =0 and solve for y getting y = 7 so (0,7) is one point on the line of Eq 1. Now let y =0 and solve for x getting x = 7/3.
So now you have the point (7/3, 0). Plot these two points and draw a line through them, extendig it as far as you like. This is the graph of Eq 1.

Now in you 2nd equation it looks like -3x2 is meant to be -3x^2 or 3x squared. If this is the case, this is a quadratic equation and is not linear. It would be a parabola.

If -3x2 means -6x then the equation is linear and would be
y= -5x-5, so you could solve this the same way Eq 1 was solved. But -3x2 is a very strange way to write -6x and so I have serius doubts as to that interpretation.

Answer 3

In general, the easiest way to do this would be to create axes on graph paper (one line horizontal, one line vertical crossing at 0,0).

Then, for values of X like -3, -2, -1, 0, 1, 2, 3, compute f(x) from the formula:
Q1 f(-3) = 9 + 7 = 16
Q1 f(0) = 7
Q1 f(2) = -6 + 7 = 1
This generates the points (-3, 16), (0, 7), (2, 1)... plot these points (they should form a straight line), and there ya go.

For Q2, this is more tricky, since it's NOT a linear equation, in that it doesn't represent a line, but a parabola.

Again, you can plug a few values for X in, but I'd do MORE values, you should see the curve come into existence as you do it. it's won't be perfect, but it'll do.

Technically, a parabola can be drawn by computing it's focus and asymptote from the formula, then drawing a curve where every point is equidistant from both the focus and the closest point on the asymptote.

Answer 4

For Q1,
f(x)= -3x +7 compare to y=mx + c,
y=f(x), m (the gradient)= -3, c (the y-intersect)=7
so the graph should has the shape like this, \

it is quite simple.
you can try substitute f(x)=0 into the equation,
f(x) = -3x + 7
0 = -3x+7
-7=-3x
x=7/3
so, when f(x)=0, x=7/3
then you substitue x=0 into the equation,
f(x)=-3x+7
f(x)=-3(0)+7
f(x)=7
so, when x=0, f(x)=7
plot these 2 points and join together with a straight line. it is a linear graph as the power of x is 1.

Answer 5

i'm not sure but i think you have to use y=mx+b equation and that one will probably be a porabala

Answer 6

Use this website.
It's simple and it will show a graph!!