#y = 2x^2 + 3x - 8#
This equation is written in standard form, or #y = ax^2 + bx + c#
To find the #x#-value of the vertex, or the axis of symmetry, we use the formula: #x = -b/(2a)#.
We know that #a = 2# and #b = 3#, so we can plug in these values into the formula and solve:
#x = -3/(2(2))#
#x = -3/4#
#------------------#
Now, to find the #y#-value of the vertex, we just plug in the value of #x# back into the original equation:
#y = 2x^2 + 3x - 8#
#y = 2(-3/4)^2 + 3(-3/4) - 8#
And now we simplify...
#y = 2(9/16) - 9/4 - 8#
#y = cancel(2)color(red)1(9/(cancel(16)color(red)8)) - 9/4 - 8#
#y = 9/8 - 9/4 - 8#
Make both fractions have the same denominator so you can subtract them:
#y = 9/8 - 18/8 - 8#
#y = -9/8-8#
Convert to mixed fraction form:
#y = -1 1/8 - 8#
#y = -9 1/8#
#------------------#
Finally, the vertex is #(-3/4, -9 1/8)#.
Hope this helps!
