y < 2x² - 4x - 3
I got the right answers for x-intercepts and the vertex...
where do you shade it? the inside or the outside of the parabola lines? how do you check it?
2007-12-01 16:59:39 · 7 answers · asked by Juan C 6 in Science & Mathematics ➔ Mathematics
Consider f(x) = 2x² - 4x - 3
f ` (x) = 4x - 4 = 0 for turning point
x = 1 for turning point
(1 , - 5) is a turning point
f "(x) = 4 (+ ve)
Thus (1 , - 5) is a MINIMUM turning point.
Cuts y axis at (0 , - 3)
Cuts x axis at :-
2x² - 4x - 3 = 0
x = 2.58 , x = - 0.58
Curve y = 2x² - 4x - 3 may now be drawn using a DOTTED line that passes thro`points:-
(2.58,0) , (- 0.58,0) , (1 , - 5) , (0 , -3)
Required region is then a shaded region that lies BELOW this dotted line.
2007-12-01 23:01:34 · answer #1 · answered by Como 7 · 5⤊ 0⤋
The parabola opens up (The leading coefficient of x^2 is pos.) The vertex is (2, -11).
The parabola comes from graphing y EQUALS to 2x² - 4x - 3
All pts (x,y) left and right and below the parabola will satisfy your inequality. Test some of them
2007-12-02 01:12:08 · answer #2 · answered by answerING 6 · 0⤊ 1⤋
dotted line on the inide, u check it by picking a point not on the line, if u pick a point above the line and plug it in for x and y in the eqaution and it is true then u shade above, if it is false then u shade below, same goes for if u had picked a point below the line, if it was true u shade below if it was false u shade above good luck!
2007-12-02 01:04:20 · answer #3 · answered by Airica ♥'s...not u! 4 · 0⤊ 1⤋
dashed line, inside, since y
2007-12-02 01:01:09
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answer #4
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answered by catsfanj 3
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No I Cannot make the help
2007-12-02 01:01:15 · answer #5 · answered by essmeluke 3 · 0⤊ 1⤋
i got no idea
2007-12-02 01:01:20 · answer #6 · answered by polaca 2 · 0⤊ 1⤋
what
2007-12-02 01:02:22 · answer #7 · answered by Anonymous · 0⤊ 0⤋