What is the the vertex of #y =5x^2+14x-6 #?

1 Answer
Mar 7, 2018

The vertex is #(-7/5,-79/5)##=##(-1.4,-15.8)#

Explanation:

#y=5x^2+14x-6# is a quadratic equation in standard form:

#y=ax^2+bx+c,#

where:

#a=5,# #b=14,# #c=-6#

The vertex is the minimum or maximum point on a parabola. To find the vertex of a quadratic equation in standard form, determine the axis of symmetry, which will be the #x#-value of the vertex.

Axis of symmetry: vertical line that divides the parabola into two equal halves. The formula for the axis of symmetry for a quadratic equation in standard form is:

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

Plug in the known values and solve for #x#.

#x=(-14)/(2*5)#

Simplify.

#x=(-14)/(10)#

Reduce.

#x=-7/5=-1.4#

To find the #y#-value of the vertex, subsitute #-7/5# for #x# and solve for #y#.

#y=5(-7/5)^2+14(-7/5)-6#

Simplify.

#y=5(49/25)-98/5-6#

Simplify.

#y=245/25-98/5-6#

Reduce #245/25# by dividing the numerator and denominator by #5#.

#y=((245-:5)/(25-:5))-98/5-6#

Simplify.j

#y=49/5-98/5-6#

In order to add or subtract fractions, they must have a common denominator, called the least common denominator (LCD). In this case, the LCD is #5#. Recall that a whole number has a denominator of #1#, so #6=6/1#.

Multiply #98/5# and #6/1# by a fractional form of #1# that will give them the LCD of #5#. An example of a fractional form of #1# is #3/3=1#. This changes the numbers, but not the values of the fractions.

#y=49/5-98/5-6xxcolor(magenta)5/color(magenta)5#

Simplify.

#y=49/5-98/5-30/5#

Simplify.

#y=(49-98-30)/5#

#y=-79/5=-15.8#

The vertex is #(-7/5,-79/5)##=##(-1.4,-15.8)#

graph{y=5x^2+14x-6 [-14.36, 14.11, -20.68, -6.44]}