A quadratic function is given. f(x) = 2x2 + 6x (a) Express the quadratic functio

Algebra
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A quadratic function is given.
f(x) = 2x2 + 6x
(a) Express the quadratic function in standard form. 
f(x) =

(b) Find its vertex and its x- and y-intercept(s). (If an answer does not exist, enter DNE.)
vertex    (xy) =
 
 
x-intercepts    (xy) =
 
 
 (smaller x-value)
(xy) =
 
 
 (larger x-value)
y-intercept    (xy) =
 
 

(c) Sketch its graph.

Jul 6th, 2015

Thank you for the opportunity to help you with your question!

The given quadratic equation is f(x) = 2x^2 + 6x

To find the standard form, factorize the equation as 

x(2x+6)=0

Now, use the MPZ (Multiplication Property of Zero) to say that

x=0 or 2x+6=0

Which gives you two solutions:

x=0 or x=−3

To find the x-intercepts, we need to solve equation 2x^2+6x=0

y - intercept is point:Y(0,0)

To compute y - coordinate of y - intercept, we will find f(0). In this example we have:

f(0)=2⋅02+6⋅0+=0

Vertex is point: V=(32,−92)

Explanation: to find the x - coordinate of the vertex we use formula

x=−b/2a

In this example: a=2,b=6,c=0. So, the x-coordinate of the vertex is:

x=−b/2a=−62⋅2=−32

f(32)=2(32)2+6⋅(32)+0=−92

Focus is point: F=(32,−358)

Explanation:

The x - coordinate of the focus is x=−b/2a

The y - coordinate of the focus is y=1−b^2/4a  +C

f(x)=2x^2+6x


Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 6th, 2015

Can you publish answers more clearly

Jul 6th, 2015

Graph.docx 

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Jul 6th, 2015

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