What are the intercepts for #Y=-2x^2-8x+6#?

1 Answer
Sep 23, 2015

Answer:

Intercepts on x-axis: #((-2 + sqrt( 7)), 0)# and #((-2 - sqrt( 7)), 0)#

Intercept on y-axis: #(0, 6)#

Explanation:

The function crosses the x-axis when #Y = 0#

Hence, to find the intercepts with the x-axis, it is necessary to find values of x satisfying

#-2 x^2 - 8 x + 6 = 0#

Which may be solved by factorisation (if there is a neat integer solution), by "completing the square", or (equivalently) by using the quadratic formula (if it does not easily factorise).

Noting that all of the coefficients are even and that #0/(-2) = 0#, the expression may be simplified, for the purpose of the attempted factorisation, by division throughout by "-2" (which also removes the inconvenient minus sign from the term in #x^2#). That is, required values of #x# will also satisfy (note the change of signs):

#x^2 + 4 x - 3 =0#

Noting that the constant "-3" has factors "3" and "1", one of which must be negative, and that the sum of these factors will be 2 or -2 (according to which is set to negative), it seems that this expression cannot be neatly factorised. Hence it is expedient to use the quadratic formula (don't waste time searching for factors if it does not look hopeful).

Using the quadratic formula, values of x satisfying the original equation are given by

#(-(-8) +- sqrt((-8)^2 - 4 (-2) (6)))/(2 (-2))#

#= (8 +- sqrt(64 +48))/(-4)#

#= (8 +- sqrt(112))/(-4)#

#= (8 +- sqrt(2 * 2 * 2 * 2 * 7))/(-4)#

#= (8 +- sqrt(4^2 * 7))/(-4)#

#= (8 +- 4 sqrt( 7))/(-4)#

#= -2 +- sqrt( 7)#

("plus or minus" became "minus or plus", but this amounts to the same thing in this particular instance as order is not important)

So the intercepts on the x-axis (recalling that #Y = 0#) are:

#((-2 + sqrt( 7)), 0)#

and

#((-2 - sqrt( 7)), 0)#

The function will cross the y-axis when #x = 0#. This may be found by simple substitution, that is

#Y = -2 (0^2) - 8 (0) + 6 = 6#

So the intercept on the y-axis is:

#(0, 6)#