How do you find the focus, directrix and sketch #y=2x^2+x-2#?

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Answer 1 · 2017-05-13T03:43:34

The focus is #(-1/4,-2)#
The equation of the directrix is #y = -9/4#

Given the standard form, #y = ax^2+bx+c#

The x coordinate of the vertex is:

#h = -b/(2a)#

This is, also, the x coordinate of the focus.

The y coordinate of the vertex is:

#k = a(h)^2+b(h)+c#

The vertical distance from the vertex to focus is:

#f = 1/(4a)#

The y coordinate of the focus is:

#y_"focus" = k+f#

The equation of the directrix is:

#y = k - f#

For the given equation, #y = 2x^2+x-2#, #a = 2#, #b=1#, and #c=-2#

Substitute these values into the above equations:

#h = -1/(2(2))#

#h = -1/4#

#k = 2(-1/4)^2+ -1/4 - 2#

#k = -17/8#

#f = 1/(4(2)#

#f = 1/8#

#y_"focus"=-17/8+1/8#

#y_"focus"= -2#

The focus is #(-1/4,-2)#

#y = -17/8-1/8 = -18/8=-9/4#

The equation of the directrix is #y = -9/4#

Here is a graph of the function:
graph{2x^2+x-2 [-10, 10, -5, 5]}