How do you find the VERTEX of a parabola #f(x)= -x^2 + 3x + 10#?

Question

Eva Abramson · Accepted Answer

Find vertex of# f(x) = -x^2 + 3x + 10#
Vertex #(3/2, 49/4)# Vertex's x coordinate is #x = (-b/2a) = -3/-2 = 3/2#. Vertex's y coordinate is #y = f(3/2) = - 9/4 + 9/2 + 10 = 49/4#.

Avery Alexander · Answer

#x=3/2 or 1 1/2#

#y=49/4 or 12 1/4#

The vertex is #(3/2, 49/4)# or #(1 1/2, 12 1/4)#.

#f(x)=-x^2+3x+10# Change #f(x)# to #y#. The formula for #y=-x^2+3x+10# is #ax^2+bx+c#, where #a=-1, #b=3, and #c=10#. Use the formula #x=(-b)/(2a)# to determine the value of #x#. #x=(-3)/(2*-1)# #x=(-3)/(-2)# #x=3/2=1 1/2# The value of #y# can be obtained by replacing #x# in the formula #y=x^2+3x+10# with #3/2#. (3/2)^2 + 3(3/2) + 10# = #y #y=-9/4+9/2+10# Multiply the terms on the right side by the fraction so that each term has a denominator of #4#. The common denominator is #4#. y = 9/4 + 9/2 * 2/2 + 10 * 4/4# #y= (18/4+40/4+9/4#) #y=49/4=12 1/4#.

Abigail Allen · Answer

undefined To find the vertex of a parabola in the form $ f(x) = ax^2 + bx + c $, you can use the formula $ x = \frac{-b}{2a} $ to find the x-coordinate of the vertex, and then substitute this value into the equation to find the corresponding y-coordinate. For the given equation $ f(x) = -x^2 + 3x + 10 $, the x-coordinate of the vertex is $ x = \frac{-3}{2(-1)} = \frac{3}{2} $. Substituting $ x = \frac{3}{2} $ into the equation gives $ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) + 10 = \frac{7}{4} $. So, the vertex of the parabola is $ \left(\frac{3}{2}, \frac{7}{4}\right) $.