Where is the vertex in the parabola y = x^2 +2x - 5? i dont understand this i need x and y intercepts and please show work?

2 Answers
May 7, 2018

Vertex (-1, -6)

Explanation:

y = x^2 + 2x - 5
The x-coordinate of vertex is given by the formula:
x = - b/(2a) = - 2/2 = - 1
The y- coordinate of Vertex is given by y(-1), when x = -1 -->
y(-1) = (-1)^2 + 2(-1) - 5 = - 6
Vertex (-1, -6)
graph{x^2 + 2x - 5 [-20, 20, -10, 10]}

May 7, 2018

Vertex #(-1, -6)#
Y-intercept #(0,-5)#
x-intercept #(1.449, 0)#
x-intercept #(-3.449, 0)#

Explanation:

Given -

#y=x^2+2x-5#

Vertex is the point where the curve turns.
To find this point - first you have to calculate
for what value of #x# the curve turns. Use the formula to find that.

#x=(-b)/(2a)#

Where -
#b# is the coefficient of #x#
#a# is the coefficient of #x^2#

#x=(-2)/(2xx1)=(-2)/2=-1#

When #x# takes the value #-1# the curve turns. At that point #x# co ordinate is #-1#, then what is #y# coordinate. Plug in the #x=-1# in the given equation.

#y=(-1)^2+2(-1)-5=1-2-5=-6#

At point #(-1,-6) # the curve turns. This point is vertex.

Vertex #(-1, -6)#
Look at the graph.

enter image source here

What is #y# intercept?
It is the point at which the curve cuts the Y-axis. Look at the graph. At #(0, -5)# the curve cuts the Y-axis.
How to find it out?
Find the What is the value of #y# when #x# takes the value #0#

At #x=0; y=0^2+2(0)-5=0+0-5=-5#

At point #(0,-5)# the curve cuts the Y-axis.
Y-intercept #(0,-5)#

What is X-intercept?

It is the point at which the curve cuts the x-axis. Look at this graph. The curve cuts the x axis at two points. Then, how to find it out. Find the value(s) of #x# when #y=0#

#x^2+2x-5=0#
Solve to find the value of #x# [Squaring method is used]

#x^2+2x=5#
#x^2+2x+1=5+1=6#
#(x+1)^2=6#
#x+1=+-sqrt6=+-2.449#
#x=2.449-1=1.449#

x-intercept #(1.449, 0)#

#x=-2.449-1=-3.449#

x-intercept #(-3.449, 0)#