# graph function

• Sep 8th 2009, 03:09 AM
mark
graph function
the first question i have is: the function of f is defined by $f(x) = (1 - 2x)(1 + 2x)$. i know how to find where it crosses the x axis, but what method do i use to find the interception of the y axis?

the second question is (note the info in the first question): the graph of y = f(x) is translated by one unit in the positive x-direction and 4 units in the positive y-direction to give the graph of y = g(x). find an expression for g(x) in the form $ax^2 + bx + c$, where a, b and c are integers.

how would i go about tackling these questions? with particular attention to the second one if possible. thankyou,
mark

ps- the answer to the second is $-4x^2 + 8x + 1$
• Sep 8th 2009, 04:09 AM
Craka
To find where it cross the x-axis, you would have made y=0 to get your two corresponding points that the graph crossed the x-axis. To find where the function crosses the y-axis you do a similar thing and sub x=0 in and find y
• Sep 8th 2009, 04:58 AM
mark
any help with the second question?
• Sep 8th 2009, 05:26 AM
mr fantastic
Quote:

Originally Posted by mark
[snip]
the second question is (note the info in the first question): the graph of y = f(x) is translated by one unit in the positive x-direction and 4 units in the positive y-direction to give the graph of y = g(x). find an expression for g(x) in the form $ax^2 + bx + c$, where a, b and c are integers.

how would i go about tackling these questions? with particular attention to the second one if possible. thankyou,
mark

ps- the answer to the second is $-4x^2 + 8x + 1$

$f(x) = (1 - 2x)(1 + 2x) = 1 - 4x^2$.

$f(x) \rightarrow g(x) = f(x - 1) = 1 - 4(x - 1)^2 = 1 - 4x^2 + 8x - 4 = -4x^2 + 8x - 3$.

$g(x) \rightarrow h(x) = g(x) + 4 = -4x^2 + 8x - 3 + 4 = -4x^2 + 8x + 1$.