Question #e5f61

1 Answer
Dec 20, 2017

Plot the solutions at #x=1,3#; plot the vertex at #(2,1)#; plot the y-intercept at #y=-3#, and you've got the graph!
graph{-x^2+4x-3 [-10, 10, -5, 5]}

Explanation:

You can find the solutions with the quadratic formula: #(-4+-sqrt(4^2-4(-1*-3)))/(2*-1)=2+--1=1,3#
The #y#-intercept is easy to find by putting #0# in for #x#, so the #y#-intercept is at #(0,-3)#.

Finally, plot the vertex by solving for the #x#-coordinate with #x=-b/(2a)=(-4)/(2*-1)=2#. Substitute #2# for #x# in the original equation to get the #y#-coordinate of the vertex: #-2^2+4(2)-3=1# So the vertex is at #(2,1)#.

If you need more points, make a table of values using numbers near the x-coordinate of the vertex.
Hope that helps!