Question

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

Answer 1

`x=-5" and maximum value "=32`

Explanation:

`"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]}

Answer 2

Axis of symmetry: `x+5=0`

Maximum value `=32`

Explanation:

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`