10Aquiz1

# 10Aquiz1 - Problem 3(3 points Solve for x 4 x-4 = e ln x 2...

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MATH 10A Quiz 1 Name (Section B0 ) Jan 14 2011 Problem 1 (2 points): Find the slope and y -intercept of the line 6 y +12 x - 2 = 0 . Solution: Rearrange the equation into slope-intercept form: y = - 12 6 x + 2 6 . Therefore, the slope is - 2, and the y -intercept is 1 3 . Problem 2 (5 points): Let f ( x ) = x 2 . Sketch the graph of the function g ( x ) = - f ( x + 1) + 4 , label and give the coordinates of the vertex of the parabola, the y -intercept, and x -intercepts. Solution: To get g ( x ), ﬁrst shift f to the left by 1 unit, reﬂect in the x-axis, then shift vertically up by 4 units to get .................................................................................................................................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . - - - - - - - - | | | | | | | | ( - 1 , 4) ( - 3 , 0) (1 , 0) (0 , 3)
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Unformatted text preview: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problem 3 (3 points): Solve for x : 4 x-4 = e ln ( x 2 ) . Solution : 4 x-4 = e ln x 2 = x 2 ( e x and ln x are inverses of each other) x 2-4 x + 4 = 0 ( x-2) 2 = 0 (factor the quadratic equation) , therefore, x = 2....
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