Question

The value of \[p\] such that the vertex of \[y = {x^2} + 2px + 13\] is \[4\] units above the \[x - axis\] is ?
A. \[ \pm 2\]
B. \[4\]
C. \[ \pm 3\]
D. \[5\]

Answer 1

Hint: The given equation is an example of a parabola . So , here the vertex is addressing the vertex of a parabola . The vertex of a parabola is the point of intersection of the parabola and its line of symmetry . The vertex of regular parabola is situated at the origin , but here it is given as \[4\] , therefore we will compare the term after simplifying the given equation to standard form .

Complete step-by-step answer:
Given : \[y = {x^2} + 2px + 13\]
The standard form of a parabola is given by :
\[{\left( {x - h} \right)^2} = a\left( {y - k} \right)\]
Here \[\left( {h,k} \right)\] represents the vertex of the parabola and \[a\] is any constant .
Now we will convert the given equation into standard form , we have
\[y = {x^2} + 2px + 13\]
Using completing the square method , we will add and subtract \[{p^2}\] , we get
\[y = {x^2} + 2px + {p^2} - {p^2} + 13\]
On solving we get ,
\[y = {\left( {x + p} \right)^2} + 13 - {p^2}\]
On simplifying we get
\[y - \left( {13 - {p^2}} \right) = {\left( {x - \left( { - p} \right)} \right)^2}\]
Now comparing the above equation with standard form of the parabola we get ,
\[\left( {h,k} \right) = \left( { - p,13 - {p^2}} \right)\] .
Now it is given that the vertex is \[4\] units above the \[x - axis\] , therefore it is the value of \[y\] coordinate .
Now equating coordinates of the \[y - axis\] we get ,
\[13 - {p^2} = 4\]
On simplifying we get ,
\[{p^2} = 9\]
Now taking square root on both sides we get ,
\[p = \pm 3\] .
Therefore , option (c) is the correct answer .
So, the correct answer is “Option c”.

Note: In the question it is given that the vertex is \[4\] units above the \[x - axis\] , which means that it is talking about the \[y - axis\] , do not compare the terms of the \[x - axis\] . Moreover , in the equation if only one of the terms is of degree \[2\] , then it resembles a parabola .