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

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Answer 1 · 2017-07-17T16:41:43

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

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

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
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$color(blue)("Determine the vertex")$

$x_("vertex")$ 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 $c=5$

Tony B Tony B

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