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

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Bunuel
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The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the [#permalink] New post   16 Nov 2022, 10:05
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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
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Re: The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the [#permalink] New post   16 Nov 2022, 20:03
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Bunuel wrote:
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



\(y = (x^2 + 2x + 1) + d\)
\(y=(x+1)^2+d\)
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.


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The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the [#permalink] New post   17 Nov 2022, 01:01
Bunuel wrote:
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


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
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Re: The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the [#permalink] New post   17 Nov 2022, 04:59
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Bunuel wrote:
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


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
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Re: The graph of y = (x^2 + 2x + 1) + d, where d > 0, intercepts with the [#permalink]
17 Nov 2022, 04:59
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