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February 10, 2016
Posted by **y912f** on Friday, November 14, 2008 at 11:25am.

in this equation

f(x)=-2x^2+4x+6

i know that you will plug in 1+a anywhere there is x. but after that i get confused.

also can you tell me how to find the domain and range of a relation

for ex. 2x+5y=10

thanks for your help!

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**Reiny**, Friday, November 14, 2008 at 11:30amf(x)=-2x^2+4x+6

f(1+a) = -2(1+a)^2 + 4(1+a) + 6

= -2(1 + 2a + a^2) + 4 + 4a + 6

= -2 -4a -2a^2 + 10 + 4a

= -2a^2 + 8

for 2x+5y=10 both the domain and range are the real numbers

You are looking at a slanted line, (slope=-2/5), so you will find the graph for whatever x you use, and for whatever y you use.

This is true for all straight line graphs, which are neither vertical nor horizontal.

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**y912f**, Friday, November 14, 2008 at 11:37amok,thanks alot.. what about this one : y=2x^2-5

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**Reiny**, Friday, November 14, 2008 at 12:04pmthis is a vertical parabola, meaning it opens either upwards or downwards

the domain of any vertical parabola is the set of real numbers

for the range:

if the parabola opens up, then y ≥ the y value of the vertex

if the parabola opens down, then y ≤ the y value of the vertex

in this case

Domain : x any real number

Range : y ≥ -5, y being real

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**y912f**, Friday, November 14, 2008 at 12:33pmok. so do u mean that to figure this out u always have to graph the relation?

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**Reiny**, Friday, November 14, 2008 at 12:45pmno, you just have to find the vertex.

Here is a quick way to find the vertex:

if y = ax^2 + bx + c

then the x of the vertex is -b/(2a)

sub that back into the function to find the y value.

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**y912f**, Saturday, November 15, 2008 at 11:10amim sry im still kinda confused. can you show me once again please and explain everything you do thnks:

y=2x+3

y=4/square root of x^2+81

thnks alot!!

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**Reiny**, Friday, November 14, 2008 at 11:30ambtw, why are you calling this "trig" ?

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**y912f**, Friday, November 14, 2008 at 11:38amits in the first chapter of my trig course. i know its review frm geometry and alg

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