You will have to use "completing the square".
y = 2x^2-8x+11
= 2(x^2 - 4x+ .... ) + 11 <---- took out the coefficient of the square term from the first two terms
= 2(x^2 - 4x + 4 - 4) <--- took half of the coefficient of the x term, then squared it, added it and immediately subtracted it again, the net result was adding zero.
= 2( (x-2)^2 - 4) + 11 <----- formed the perfect square plus the left over constant
= 2(x-2)^2 - 8 + 11 <---- dististributed the 2 over the perfect square and the left-over constant
= 2(x-2)^2 + 3 <---- added the stuff at the end
all done, now I can read the vertex
which is .......
and if I let x = 0 in the original (or this last) equation, I get
y = 11
So plot the vertex, plot the y-intercept and form a neat parabola from the vertex to hit (0,11). Complete the other half of the parabola.
It should look like this:
http://www.wolframalpha.com/input/?i=plot+y+%3D+2x%5E2-8x%2B11,+y+%3D+2(x-2)%5E2+%2B+3
notice I graphed both forms of the equation and the graphs coincide. This was a neat way to check my work.
write the equation in standard form and then graph the equation.
y=2x^2-8x+11
1 answer
1 answer