### Video Transcript

Find the vertex of the graph of π¦
equals five multiplied by π₯ plus one all squared plus six.

Here, we have a quadratic
function. And itβs been expressed in its
completed square or vertex form. In general, this form is π
multiplied by π₯ plus π all squared plus π. So, comparing the general form with
our quadratic, we see that the value of π is five, the value of π is one, and the
value of π is six. We were asked to use this form to
find the vertex or turning point of the graph of this quadratic function.

Now, we should recall that all
quadratic functions have the same general shape, which is called a parabola. When the value of π is positive,
the parabola will curve upwards. And when the value of π is
negative, the parabola will curve downwards. The vertex of the graph is its
minimum point in the first instance and its maximum point in the second. Our value of π in this question is
five, which is positive. So, weβre in this first instance
here. Weβre looking for the coordinates
of the minimum point of this graph. Letβs consider then how we can
minimize this quadratic function and for what value of π₯ this minimum will
occur. And then, weβll see how we can
generalize.

Letβs consider each part of the
function in turn, starting with π₯ plus one squared. Well, this is a square. And we know that squares are always
nonnegative. Whatever value of π₯ we choose,
once we have added one and then squared it, the result will be zero or above. So, the minimum value of π₯ plus
one all squared is zero. We then multiply this by five. But, of course, anything multiplied
by zero is still zero. So, the minimum value of five
multiplied by π₯ plus one all squared is still zero. We then add six, so the entire
function increases by six units, which means that its minimum value overall is
six. This gives the minimum value of the
function or the minimum value of π¦. So, it is the π¦-coordinate of the
vertex.

We then need to determine what
value of π₯ causes this to be the case. Well, itβs the value of π₯ such
that π₯ plus one all squared takes its minimum value of zero. Itβs, therefore, the value of π₯
such that the expression inside the parentheses, thatβs π₯ plus one, is equal to
zero. The solution to this simple linear
equation is π₯ equals negative one. So, we find that the π₯-coordinate
of the vertex is negative one. The coordinates of the vertex of
this graph then are negative one, six.