How do you find the axis of symmetry, and the maximum or minimum value of the function #y= -x^2-10x+7#?

#"We require to find the vertex and determine if maximum"#
#"or minimum turning point"#

#"the equation of a parabola in "color(blue)"vertex form"# is.

#•color(white)(x)y=a(x-h)^2+k#

#"where "(h,k)" are the coordinates of the vertex and a"#
#"is a multiplier"#

#x=h" is the axis of symmetry"#

#"to obtain this form use "color(blue)"completing the square"#

#y=-(x^2+10x-7)#

#color(white)(y)=-(x^2+2(5)x+25-25-7)#

#color(white)(y)=-(x+5)^2+32#

#color(magenta)"vertex "=(-5,32)#

#"Since "a<0" then maximum turning point "nnn#

#"axis of symmetry is "x=-5#

#"and maximum value "=32#
graph{-x^2-10x+7 [-80, 80, -40, 40]}

The given equation:

#y=-x^2-10x+7#

#y=-(x^2+10x+25)+25+7#

#y=-(x+5)^2+32#

#(x+5)^2=-(y-32)#

The above equation is in standard form of downward parabola #(x-x_1)^2=-4a(y-y_1)# which has

Axis of symmetry: #x-x_1=0#

#x+5=0#

Vertex #(x-x_1=0, y-y_1=0)#

#(x+5=0, y-32=0)\equiv (-5, 32)#

Maximum value of given quadratic function will be at the vertex #(-5, 32)# of given downward parabola: #y=-x^2-10x+7#

Hence, the maximum value of given function is obtained by setting #x=-5# in the function as follows

#y(-5)=-(-5)^2-10(-5)+7#

#=32#