How do you find the vertex of a parabola f(x)=x^2 +10x-11f(x)=x2+10x−11?

1 Answer
Sridhar V.
Mar 23, 2018

Vertex of the parabola is at :color(blue)((-5, -36)(−5,−36)

Explanation:

The Standard form for the quadratic equation is :

y=f(x)=ax^2+bx+c = 0 y=f(x)=ax2+bx+c=0

Given:

y=f(x) = x^2+10x-11y=f(x)=x2+10x−11

where a=1; b=10 and c= -11a=1;b=10andc=−11

Find the x-coordinate of the vertex using the formula -b/(2a)−b2a

x=-10/(2*1)x=−102⋅1

rArr x=-5⇒x=−5

To find the y-coordinate value of the vertex, set x=-15x=−15 in y= x^2+10x-11y=x2+10x−11

y=(-5)^2+10(-5)-11y=(−5)2+10(−5)−11

Simplify the right-hand side.

rArr 25-50-11⇒25−50−11

rArr 25-61⇒25−61

rArr -36⇒−36

y=-36y=−36

Hence,

Vertex of the parabola is at :color(blue)((-5, -36)(−5,−36)

Also observe that, since the coefficient of x^2x2 term (a)(a) is less than zero, the parabola opens down and the graph has a minimum at the vertex.

Please refer to the image of the graph below:

enter image source here

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