The graph of y = (x^2 + 2x + 1) + d, where d 0, intercepts with the

Question

The graph of  y = (x^2 + 2x + 1) + d , where d > 0, intercepts with the x-axis at how many points?

A. 0
B. 1
C. 2
D. 3
E. 4

chetan2u · Answer

y = (x^2 + 2x + 1) + d 
 y=(x+1)^2+d 
So, lowest value of y will be when x=-1, and y will be d then.

The equation  y = (x^2 + 2x + 1) + d  is a quadratic equation and will cut the x-axis at just two points or no points. 
Since, the lowest value of y is d, which is greater than 0, the quadratic equation does not intersect x axis.

A

gmatophobia · Answer

To find the points of intersection with x axis, put y = 0

0 = x^2 + 2x + (d+1)

The roots of this equation are the points of intersection. However, if the roots are not real, then the graph doesn't intersect x axis.

Roots =  \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

=  \frac{-2 \pm \sqrt{4 - 4(d+1)}}{2a}

We know that d is > 0, hence 4(d+1) > 4

Thus the value  \sqrt{4 - 4(d+1)}  will not yield any real roots.

Hence, we can conclude that the graph does not cut x-axis.

Option A

SaquibHGMATWhiz · Answer

Solution:

We have  y = (x^2 + 2x + 1) + d  
 We are asked the number of x-intercepts  
 For which we have to plug y = 0  
 The equation thus becomes  (x^2+2x+1)+d=0 
 ⇒(x+1)^2+d=0   
 ⇒(x+1)^2=-d  
 This is obviously not possible because square of a number x + 1 can never be negative i.e., -d

Hence the right answer is    Option A

The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the

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