Find the axis of symmetry, Vertex and Y- Intercept and sketch the Parabola and label all parts? #y= x^2 + 6x +4#

i dont understand any of it

1 Answer
Apr 13, 2018

Talking about what the equation is saying.

Explanation:

The general shape of a quadratic graph is called a parabola. It is just a name but one worth remembering. On the whole we have the generic shapes #nn and uu #. If you ever do higher maths you will come across #sub and sup# as well.

#y=ax^2 -># where a is positive we have the general shape #uu#
#y=ax^2-> # where a is negative we have the general shape #nn#

If #a# is some where between but not including #-1" to "+1#
then the shape #nn or uu# is widened

If #a # is greater than 1 or less than -1 the general shape of #nn or uu# is narrowed.

Note that #y=x^2# is of form #y=ax+0x+0#

This part in the given question is #y=1x^2+...." so "a=1#
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What happens if we add the constant 4

#y=x^2+4#

The 4 is the value on the y-axis through which the graph passes. It is the y-intercept. So it lifts #y=x^2# upwards by 4.
No matter how you change the equation, if the constant is 4 then the graph will always pass through that point on the y-axis.
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What happens if we add #6x#

This particular part defines the #x# value of the vertex

#x_("vertex")=(-1/2)xx6 = -3#

By substitution you can determine the value of #y_("vertex")#

There are other ways of determining the vertex.
Completing the square or half way between where the graph crosses the x-axis.

#color(brown)("This will do for now! You will learn more as you go on.")#

Tony B