How do you find and classify all the critical points and then use the second derivative to check your results given #y=x^2+10x-11#?

Question

Ezekiel Alexander · Accepted Answer

Vertex #(-5, -36)#
Y-intercept #(0, -11)#
X-intercepts #(-11, 0) #and #(1, 0)#

Given -

#y=x^2+10x-11#

It is a quadratic equation .
It has only one critical point.
It is the vertex.

#x=(-b)/(2a)=(-10)/(2 xx 1)=-5#

At #x=-5; y= (-5)^2+10(-5)-11#

#y= 25-50-11=25-61=-36#

Vertex is #(-5, -36)#

Derivatives of the function are

#dy/dx=2x+10#
#(d^2y)/(dx^2)=2 > 0#

Its second derivative is greater than zero. The curve is concave upwards.

Its other important points are

Y-intercept

At #x=0; y=0^2+10(0)-11=-11#

At #(0, -11# the curve cuts the Y-axis

X- intercepts

At #y=0; x^2+10x-11=0#

# x^2+11x-x-11=0#
#x( x+11)-1(x+11)=0#
#(x+11)(x-1)=0#
#x+11=0#
#x=-11#

#(-11, 0) # is one of the x- intercept

#x-1=0#
#x=1#

#(1, 0)# is another x-intercept.

At points #(-11, 0) #and #(1, 0)# , the curve cuts the x-axis

Ezekiel Alexander · Answer

undefined To find critical points, differentiate the function, set the derivative equal to zero, and solve for x. To classify the critical points, use the second derivative test: if the second derivative is positive at a critical point, it's a local minimum; if negative, it's a local maximum; if zero, the test is inconclusive.

1. Find the derivative: y' = 2x + 10.
2. Set y' = 0: 2x + 10 = 0, x = -5.
3. This critical point is a local minimum since the second derivative is positive at x = -5: y'' = 2.

So, the critical point is x = -5, and it's a local minimum.