How do you find the vertex of #y=2x^2+3x-8#?

Question

Eva Abramson · Accepted Answer

The vertex is #(-3/4, -9 1/8)#.

Here's how I did it:

#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!

Jameson Allen · Answer

undefined To find the vertex of the quadratic function $ y = 2x^2 + 3x - 8 $, you can use the formula for the x-coordinate of the vertex, which is $ x = \frac{-b}{2a} $, where $ a $ is the coefficient of the $ x^2 $ term and $ b $ is the coefficient of the $ x $ term. Once you have found the value of $ x $, substitute it back into the equation to find the corresponding value of $ y $. So, for $ y = 2x^2 + 3x - 8 $, $ a = 2 $ and $ b = 3 $. Plug these values into the formula to find $ x $, and then substitute $ x $ back into the equation to find $ y $.