Converting Between Standard Form and Vertex Form
Hello! Welcome to this video on converting between standard form and vertex form of a quadratic equation. Before we start converting, let’s review what these forms look like.
Standard form of a quadratic equation is:
\(y=ax^{2}+bx+c\)
Where \(a\), \(b\), and \(c\) are real numbers, and \(a\neq 0\).
Vertex form of a quadratic equation is:
\(y=a(x-h)^{2}+k\)
Where \(a\), \(h\), and \(k\) are real numbers, \(a\neq 0\), and \((h,k)\) is the vertex of the parabola.
Now that we’ve reviewed these forms, let’s start by converting a standard form equation to vertex form.
Convert the standard form equation \(y=x^{2}+14x-9\) to vertex form.
In order to do this, we need to complete the square. To complete the square, \(a\) must equal 1. In this equation, \(a\) is already 1, so we can move on to the next step.
Find the \(b\)-value, divide it by 2, and square it.
\((\frac{b}{2})^{2}=(\frac{14}{2})^{2}=(7)^{2}=49\)
Now, add and subtract this value after the \(x\)-term.
\(y=x^{2}+14x+49-49-9\)
I’m going to add parentheses around these first three terms because this will be our perfect square trinomial.
\(y=(x^{2}+14x+49)-49-9\)
This trinomial factors to the perfect square \((x+7)^{2}\).
\(y=(x+7)^{2}-49-9\)
Remember, the point of completing the square is to get it to factor into something that looks like this. So now, all we have to do is combine our like terms right here.
\(y=(x+7)^{2}-58\)
Now, notice that we went through this completing the square process fairly quickly. If you want some more help on this, see one of our other videos where we go more in depth over how to do this process.
Let’s try another problem where we convert a standard form equation to vertex form.
\(y=-2x^{2}-4x+8\)
Remember, in order to complete the square, the value of a must be 1. Factor a –2 out of the right side of the equation.
\(y=-2(x^{2}+2x-4)\)
Remember, we factor a –2 out of each of the terms. Now, identify the \(b\)-value, halve it, and square it. So, we want the \(b\)-value of this trinomial where a is 1 \((x^{2}+2x-4)\), so our \(b\)-value is 2.
\((\frac{b}{2})^{2}=(\frac{2}{2})^{2}=(1)^{2}=1\)
Add and subtract this value after the \(x\)-term.
\(y=-2(x^{2}+2x+1-1-4)\)
Put parentheses around the first three terms to identify the perfect square trinomial.
\(y=-2((x^{2}+2x+1)-1-4)\)
Factor the perfect square trinomial.
\(y=-2((x+1)^{2}-1-4)\)
Combine like terms inside the parentheses.
\(y=-2((x+1)^{2}-5)\)
Now, the final step in simplifying this just a little bit more, is distributing the –2 inside this larger set of parentheses, so to the \((x+1)^{2}\) and to the –5.
\(y=-2(x+1)^{2}+10\)
Now let’s move on to converting a vertex form equation to standard form. This conversion process is much simpler.
Convert the equation \(y=3(x-4)^{2}+7\) to standard form.
The first thing to do is expand the squared term.
\(y=3(x-4)(x-4)+7\)
And then we can FOIL to simplify this part.
\(y=3(x^{2}-4x-4x+16)+7\)
Now, we can combine like terms.
\(y=3(x^{2}-8x+16)+7\)
Now, distribute 3 to the trinomial.
\(y=3x^{2}-24x+48+7\)
Finally, combine like terms.
\(y=3x^{2}-24x+55\)
And there’s our equation in standard form.
Let’s try one more example before we go.
Convert the equation \(y=-11(x+8)2-9\) to standard form.
Start by expanding the squared binomial.
\(y=-11(x+8)(x+8)-9\)
And again, we can FOIL these two terms.
\(y=-11(x^{2}+8x+8x+64)-9\)
We can simplify by combining like terms.
\(y=-11(x^{2}+16x+64)-9\)
Then, distribute –11 to the trinomial.
\(y=-11x^{2}-176x-704-9\)
Finally, combine like terms.
\(y=-11x^{2}-176x-713\)
And there you have it!
I hope this video on converting between standard form and vertex form of a quadratic equation was helpful. Thanks for watching, and happy studying!