How do you graph #y=2x^2 -5x -3#?
2 Answers
You can easily do this by inserting values in x and plotting some points.
Explanation:
You can graph this easily by making a table of x and y values. But first, we must find the vertex
To solve for the vertex, we'll start with this formula (
Now to find
The vertex is
After plotting the vertex, just plug in other values into x and graph them. Start with simple ones such as the y-intercept (set
graph{2x^2-5x-3 [-14.24, 14.24, -7.12, 7.12]}
y= (2x+1) (x-3) graph{2x^2-5x-3 [-6.07, 7.977, -6.18, 0.847]}
Explanation:
Cross factorise the quadratic equation to get the (2
Then, put:
2
so make
And then,
so move -3 over
So
To find the
- Factorise
#2x^2# - 5#x# with the coefficient of#2x^2# which is 2 (because#x# has to have a coefficient of only 1 !) - Which makes it 2(
#x^2# - 2.5) - 3 - Then 2 [
#(x-1.25)^2# - 1.56 ) ] -3
You get 1.25 from dividing 2.5 by 2. You always have to divide the number by 2. Also, you get 1.56 (rounded up to 3 s.f) from squaring 1.25. These are all rules! - Then expand the brackets, so
2#(x-1.25)^2# -3.12-3 - 2
#(x-1.25)^2# - 6.12
So your
And your minimum point will be (1.25,-6.12).