# Video: Using Information about a Given Sketch of a Quadratic Graph to Determine the Characteristics of a Related Quadratic Graph

The vertex of the parabola in the given π₯-π¦ plane is (0, π). Which of the following is true about the parabola with the equation π¦ = βπ(π₯ + π)Β² + π? π¦ = ππ₯Β² + π. [A] The vertex is (βπ, π) and the graph opens downward. [B] The vertex is ( π, π) and the graph opens downward. [C] The vertex is (βπ, π) and the graph opens upward. [D] The vertex is (π, π) and the graph opens upward.

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### Video Transcript

The vertex of the parabola in the given π₯π¦ plane is zero, π. Which of the following is true about the parabola with the equation π¦ equals negative π times π₯ plus π all squared plus π? And weβve been given the graph of a quadratic function π¦ equals ππ₯ squared plus π. And the four options are. A) The vertex is negative π, π and the graph opens downward. B) The vertex is π, π and the graph opens downward. C) The vertex is negative π, π and the graph opens upward. And D) The vertex is π, π and the graph opens upward.

Letβs just recall what we know about quadratic equations. The general form of the quadratic equation is that π¦ is equal to some constant times π₯ squared. Letβs call that π plus some other constant. Letβs call that π times π₯ plus some other constant. Letβs call that π. Now itβs important that the value of π is not zero. Because if we had zero π₯ squared, then we wouldnβt have a quadratic equation. But either or both of constants π and π could be equal to zero. Youβd still have a quadratic equation.

Now we also know that the shape of the graph of a quadratic is a parabola. And this is either a symmetrical curve like this, which is open upwards, or a symmetrical curve like this, which is open downwards. If the value of π in our general equation is greater than zero. If itβs positive, then we have an open upwards curve. You can think of this as being positive and happy. So it looks like a smiley face. But if the value of π is negative, then you have an open downward type graph. And you can think of this if youβre negative, youβre sad. And it looks like a sad face. Also, the value of π tells us the value of the π¦-intercept. Thatβs the π¦-coordinate where it cuts the π¦-axis.

Remember, on the π¦-axis, the π₯-coordinate is zero. So this term is gonna be π times zero squared, which will be zero. This term will be π times zero, which is zero. So the π¦-coordinate will just be equal to whatever this number is, π. And just another couple of things. The turning point of the curve we call a vertex. And if itβs an open upwards curve, then that vertex will be the minimum π¦-value on the graph. And if itβs an open downwards curve, then the vertex will be at the maximum π¦-value on the graph. And you also need to remember that you can calculate the π₯-coordinate of the vertex by evaluating negative π over two π.

Okay then, letβs look at the graph that they gave us in the question. First, itβs open downward. Weβve got a negative sad curve. That tells us that the value of π must be less than zero. π is negative. Thereβs no term for π in our equation. So we know that the value of π is zero. And the value of π is, well, π. So our π¦-intercept is at zero, π. And the question also told us that the vertex of the parabola is at zero, π. So this means that our parabola is symmetrical about the π¦-axis. Now, the values of π, π, and π that weβve been given in our equation π¦ equals negative π times π₯ plus π all squared plus π. Are gonna be the same as the values of π, π, and π in the graph that we were given. So letβs do a little bit of analysis and see if we can work out what the coefficient of π₯ squared is gonna be in our new equation. And also what the π₯-coordinate of the vertex of that curve is going to be.

Well, we can start off by remembering that π₯ plus π all squared means π₯ plus π times π₯ plus π. And if we multiply each term in the first set of parentheses by each term in the second set of parentheses. And we can see that positive π₯ times positive π₯ gives us positive π₯ squared. Positive π₯ times positive π gives us positive π₯π. Or we can write that as positive ππ₯. Positive π times positive π₯ gives us positive ππ₯. And positive π times positive π is positive π squared. And weβve still got our negative π times all of that. And weβve got plus π on the end. We can add together the like terms, positive ππ₯ and positive ππ₯ to give us positive two ππ₯. Then, we can distribute negative π through the parentheses. Negative π times positive π₯ squared is negative ππ₯ squared. Negative π times positive two ππ₯ is negative two πππ₯. And negative π times positive π squared is negative ππ squared. And donβt forget weβve got positive π on the end. So π¦ is equal to negative π₯ squared minus two πππ₯ minus ππ squared plus π.

Now, this starts to look a bit confusing because weβve used π, π, and π for different things. Weβve got π, π, and π in our general form of the quadratic equation. And then, weβve got the π, π, and π that weβre using for specific values in this question. So Iβm just gonna change the π, π, and π in our general form of the quadratic equation to π, π, and π. And that makes the π₯-coordinate of the vertex negative π, thatβs the coefficient of π₯, over two times π. Thatβs the coefficient of π₯ squared. So in a rearranged equation, the coefficient of π₯ squared, our equivalent of π in the general formula, is negative π. The coefficient of π₯, our equivalent of π in the general formula, is negative two ππ. And our constant term, the equivalent of π in the general formula, is negative ππ squared plus π.

Now we know that the value of π is less than zero. So if π is negative, then negative of π must be positive. Negative π must be greater than zero. And we said that when our coefficient of π₯ squared is greater than zero, itβs positive. If itβs positive, itβs happy. If itβs happy, itβs smiling. And a smiling parabola is also known as an open upward curve. Now, we need to use this formula to work out the π₯-coordinate of our vertex. So thatβll be the negative of the π₯-coefficient divided by two times the π₯ squared coefficient. And in our case, we said that the π₯-coefficient was negative two ππ. And the π₯ squared coefficient was negative π.

Now, we can divide top and bottom by two to give us the negative of negative one ππ over one times negative π. So that simplifies to the negative of negative ππ over negative π. Well, we donβt need the parentheses on the bottom anymore. So I can divide top and bottom by negative π. And of course, negative π divided by negative π is one in each case. So this all simplifies to negative π. The π₯-coordinate of our vertex is negative π.

Now, letβs just make a little space to calculate the π¦-coordinate of the vertex. And to find out the π¦-coordinate of the vertex, we simply need to substitute in that value that we found for π₯, negative π, into our original equation. π¦ equals negative π times π₯ plus π all squared plus π. So that π¦-value is negative π times negative π plus π all squared plus π. But of course, negative π plus π is just equal to zero. So weβve got negative π times zero squared plus π. Well, zero squared is zero. And zero times negative π is just zero. So the π¦-coordinate at the vertex is just π. So we found that the parabola with the equation π¦ equals negative π times π₯ plus π all squared plus π is open upwards. And it has a vertex at negative π, π.

Option A had the correct vertex. But it said the graph was open downwards. Option B had the wrong vertex and said the graph was open downward. Option D said the graph was open upward, which was good. But it did have the wrong vertex. This leaves us with option C. In our case, the vertex is negative π, π. And the graph opens upward.