How do you find the x and y intercepts for #y=2x^2-4x+3#?

2 Answers
Alan P.
Nov 13, 2015

y-intercept #=3#
There are no x-intercepts

Explanation:

Given #y=2x^2-4x+3#

The y-intercept is the value of #y# when #x=0#
#color(white)("XXX")y=2(0)^2-4(0)+3 = 3#

For a quadratic in the general form:
#color(white)("XXX")y=ax^2+bx+c#
the determinant #Delta = b^2-4ac# indicates the number of zeros.

#Delta {(< 0, rArr, "no solutions"),(=0, rArr, "one solution"),(>0,rArr,"two solutions"):}#

In this case
#color(white)("XXX")Delta = (-4)^2-4(2)(3) < 0#
so there are no solutions (i.e. no values for which the expression is equal to zero).

This can also be seen from a graph of this equation:
graph{2x^2-4x+3 [-6.66, 13.34, -0.64, 9.36]}

Vinícius Ferraz
Nov 13, 2015

#(0, 3)#

Explanation:

#x = 0 \Rightarrow y = 0 - 0 + 3#

#y = 0 \Rightarrow x = frac{-b ± sqrt {b^2 - 4ac}}{2a}#

#a = 2, b = -4, c = 3#

But #Delta# < 0, then there is no real root #(x_0, 0)#.

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