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

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Answer 1 · 2016-07-30T02:03:40

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

Look at the graph

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