Video: Determining the Coordinates of the Vertex of the Graph of a Quadratic Equation in Vertex Form

Find the vertex of the graph of 𝑦 = 5(π‘₯ + 1)Β² + 6.


Video Transcript

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

Here, we have a quadratic function. And it’s been expressed in its completed square or vertex form. In general, this form is π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž. So, comparing the general form with our quadratic, we see that the value of π‘Ž is five, the value of 𝑝 is one, and the value of π‘ž 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‘Ž is positive, the parabola will curve upwards. And when the value of π‘Ž 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‘Ž in this question is five, which is positive. So, we’re in this first instance here. We’re looking for the coordinates of the minimum point of this graph. Let’s consider then how we can minimize this quadratic function and for what value of π‘₯ this minimum will occur. And then, we’ll see how we can generalize.

Let’s 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 𝑦. So, it is the 𝑦-coordinate of the vertex.

We then need to determine what value of π‘₯ causes this to be the case. Well, it’s the value of π‘₯ such that π‘₯ plus one all squared takes its minimum value of zero. It’s, therefore, the value of π‘₯ such that the expression inside the parentheses, that’s 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.

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