How do you find the axis of symmetry, and the maximum or minimum value of the function #y = -x^2 - 3x -5#?

Question

Nathan Axel · Accepted Answer

Axis of symmetry is # x =-1.5 #, maximum value is #-2.75#
and minimum value extends to #-oo#.

#y= -x^2-3x-5 or y= -(x^2+3x) -5 # or #y=-(x^2+3x+1.5^2)+2.25 -5 # or #y=-(x^2+1.5)^2-2.75 # . This is vertex form of equation #y=a(x-h)^2+k ; a=-1 ,h=-1.5 ,k=-2.75 # Therefore vetex is at #(h,k) or (-1.5, -2.75)# Axis of symmetry is #x= h or x =-1.5 ; a# is negative, so parabola opens downward. Therefore vertex is the maximum point #(-1.5, -2.75) :.# Maximum value is #-2.75# and minimum value extends to #-oo#. graph{-x^2-3x-5 [-20, 20, -10, 10]} [Ans]

Jace Anderson · Answer

undefined To find the axis of symmetry of the function $ y = -x^2 - 3x - 5 $, use the formula $ x = \frac{-b}{2a} $, where $ a $ and $ b $ are the coefficients of the quadratic function. Then, substitute the value of $ x $ into the function to find the corresponding $ y $ value.

For the given function:
$ a = -1 $ (coefficient of $ x^2 $)
$ b = -3 $ (coefficient of $ x $)

Now, plug these values into the formula:
$ x = \frac{-(-3)}{2(-1)} = \frac{3}{-2} = -\frac{3}{2} $

The axis of symmetry is $ x = -\frac{3}{2} $.

To find the maximum or minimum value, substitute the value of $ x $ back into the original function:
$ y = -\left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) - 5 $
$ y = -\frac{9}{4} + \frac{9}{2} - 5 $
$ y = -\frac{9}{4} + \frac{18}{4} - \frac{20}{4} $
$ y = -\frac{11}{4} $

So, the maximum value of the function is $ -\frac{11}{4} $.