let f(x)=2x^2-8x-9

Question

This is for my is for my IB math class, I really need help!!!!! Please help

Answer 1

 

f(x) = 2x² - 8x - 9       2 > 0 -----> the graph is a parabola, upward [ like a bucket ]

 

Vertex { -(-8)/2(2) ; 2(2)² - 8(2) - 9 }        ;   vertex ( 2 ; -17 ) 

 

y-intercept ( 0 ; -9 )

 

x-intercepts { [ -(-8) - √(-8)² - 4(2)(-9) ]/2(2)  ; 0 }<---> [( 8 - √136)/4 ; 0]<--->(2-√8.5 ; 0)  

 

                  { [ -(-8) + √(-8)² - 4(2)(-9) ] / 4    ; 0 }<-------------------------------->(2+√8.5 ; 0)

 

Domain f = R the set of all real numbers : (-infinity; +infinity)

 

Range f : [-17; +infinity)

Answer 2

Because it's here, I'll just complete the square on this one:

 

• f(x)=2x²-8x-9
• f(x)=2(x²-4x)-9
• f(x)=2(x²-4x+4)-9-8 (we added 8, so we now have to subtract 8)
• f(x)=2(x-2)²-17
• This gives us the vertex of (2,-17)
• 0=2(x-2)²-17
• 2(x-2)²=17
• (x-2)²=17/2
• x-2=+/- root(17/2)
• x=2+/- root(17/2)
• x=2+/- root(34)/2

 

The graph would be a typical parabola shape that opens up (2 is positive), with the vertex at 2,-17, and a y intercept of -9