Question

How do you graph the quadratic function and identify the vertex and axis of symmetry for `y=1/2x^2+4x+5`?

Answer 1

`x_("intercpts")=-4+-sqrt(6)larr" Exact value"`
`y_("intercept")=5`
Vertex `->(x,y)=(-4,-3)`

Explanation:

`color(blue)("The different way to find "x_("vertex"))`

Write as: `y=1/2(x^2+8x)+5`

Axis of symmetry `->x_("vertex")=(-1/2)xx8= -4`

`y_("vertex")=1/2(-4)^2+4(-4)+5 = -3`

Vertex`->(x,y)=(-4,-3)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("The formula method for x-intercepts")`

Given: `y=1/2x^2+4x+5`

`y=ax^2+bx+c => x=(-b+-sqrt(b^2-4ac))/(2a)`

This really is worth committing to memory.

Where `a=1/2"; "b=4"; "c=5`

`x=(-4+-sqrt(4^2-4(1/2)(5)))/(2(1/2))`

`x=-4+-sqrt(6)larr" Exact value"`

`x~~-1.55 and -6.45` to 2 decimal places
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the vertex")`

`x_("vertex")` is mid point of intercepts which is:

`((-4-sqrt(6))+(-4+sqrt(6)))/2 =-4`

by substitution `y_("vertex")=-3`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)("Determine the y-intercept")`

It is the value of the constant `c=5`