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Write an equation for a function with the following characteristics: the shape of y=...

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jlozoyadragon | Student, Undergraduate | eNotes Newbie

Posted February 23, 2010 at 10:32 AM via web

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Write an equation for a function with the following characteristics: the shape of y= square root of x, shifted left 6 units, and down 5 units.

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neela | High School Teacher | (Level 3) Valedictorian

Posted February 23, 2010 at 12:07 PM (Answer #1)

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The function is y = sqrtx Or y = x^(1/2).

If the transformation is a shift of (h, k), then the equation becomes:

X = x+h and Y = x+k and the new equation is

Y-k =(X-h)^(1/2) Or

y-k = (x-h)^(1/2). Therefore,

(y-k)^2 = (x-h) is a parabola.

The vertex is at ( h, k) or  (-6, -5)

Also  y^2-2yk+ k^2-h -x = 0 is a quadratic in x with a discriminant, 4k^2-4(k^2+h-x) = 4(x-h) .

Since the discriminant is -h or +5  as h =5. Therefore parabola intercepts Y axis at two points  k+sqrt(0+h) and k+sqrt(0+h) or at -5+sqrt6  or  -5-sqrt6

y = k  Or y = -5 or y+5=0  is the axis of symmetry.

Firther, (y-k)^2 = 4(1/4)(x-h) . 2(1/4) or 1/2 is the latus rectum of the parabola which is independent of (h,k).

When y = 0 in (y-k)^2 = x-h,  x=k^2+h Or x = (5)^2-6 = 19 is the intercept of x axis.

The focus of the parabola is 1/4 units right of h. Or the coordinates of the focus is (h+1/4, k) Or

(-6+1/4, -5) = (-5.75, -5).

Thus the  parabola ocupies all 4 quadrarnts, with its vertex at (-6,-5), focus (-5.75, -5) , axis of symmetry y +5 = 0, latus rectum 1/2, only one interceptsof x axis at 19 units and two y intercepts at -5+sqrt6 and -5+sqrt6. The parabola opens to the right and is compressed along x axis.

 

 

 

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job518 | College Teacher | (Level 2) Adjunct Educator

Posted February 23, 2010 at 12:18 PM (Answer #2)

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Starting with the parent function f(x) = sq rt x. (where sq rt is "the square root of" . I always remember that vertical shifts are "tagged on the end" and horizontal shifts are part of what is happening to x, because the x values are along the horizontal axis (x-axis). So a horizontal shift of 6 to the left, is going to be x + 6. It is the opposite of what you think here, adding moves the graph left and subtracting moves the graph right. Now, we have sq rt (x+6) => the sq rt function shifted left. We still need to move the graph down 5 units. Since this is a change in the location along the y-axis, we just subtract (because it is moved down) from the term sqrt(x+6). So we now have f(x) = sqrt(x+6) -5.  

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