Discuss and answer Exercise 135 as a group. The equation of...

Discuss and answer Exercise 135 as a group. The equation of a parabola can be determined if three points on the parabola are known. To do so, start with $y=a x^{2}+b x+c$. Then substitute the $x$ - and $y$-coordinates of the first point into the equation. This will result in an equation in $a, b$, and $c$. Repeat the procedure for the other two points. This process yields a system of three equations in three variables. Next, solve the system for $a, b$, and $c$. To determine the equation of the parabola, substitute the values determined for $a, b$, and $c$ into the equation $y=a x^{2}+b x+c$. Three points on a parabola are $(0,12),(3,-3)$, and $(-2,32)$. a) Individually, determine a system of equations in three variables that can be used to determine the equation of the parabola. Then compare your answers. If each member of the group does not have the same system, determine why. b) Individually, solve the system and determine the values of $a, b$, and $c$. Then compare your answers. c) Individually, write the equation of the parabola passing through $(0,12),(3,-3)$, and $(-2,32)$. Then compare your answers. d) Individually, write the equation in $$ y=a(x-h)^{2}+k $$ form. Then compare your answers. e) Individually, graph the equation in part d). Then compare your answers.

Discuss and answer Exercise 135 as a group.
The equation of a parabola can be determined if three points on the parabola are known. To do so, start with $y=a x^{2}+b x+c$. Then substitute the $x$ - and $y$-coordinates of the first point into the equation. This will result in an equation in $a, b$, and $c$. Repeat the procedure for the other two points. This process yields a system of three equations in three variables. Next, solve the system for $a, b$, and $c$. To determine the equation of the parabola, substitute the values determined for $a, b$, and $c$ into the equation $y=a x^{2}+b x+c$.
Three points on a parabola are $(0,12),(3,-3)$, and $(-2,32)$.
a) Individually, determine a system of equations in three variables that can be used to determine the equation of the parabola. Then compare your answers. If each member of the group does not have the same system, determine why.
b) Individually, solve the system and determine the values of $a, b$, and $c$. Then compare your answers.
c) Individually, write the equation of the parabola passing through $(0,12),(3,-3)$, and $(-2,32)$. Then compare your answers.
d) Individually, write the equation in
$$
y=a(x-h)^{2}+k
$$
form. Then compare your answers.
e) Individually, graph the equation in part d). Then compare your answers.

Key Concepts

Converting to Vertex Form

Converting the standard form of a quadratic function to vertex form, which is written as y = a(x - h)² + k, enables a clearer understanding of the graph's geometry, including its vertex and direction of opening. This process often involves techniques such as completing the square and is an important concept in analyzing and graphing quadratic functions.

Solving Systems of Equations

The process of solving systems of equations involves finding the values of variables that satisfy all equations simultaneously. Techniques such as substitution, elimination, or using matrices are crucial in algebra, and mastering these methods enables one to solve for coefficients in models like quadratic functions derived from point data.

Graphing Parabolas

Graphing a parabola involves plotting key features such as the vertex, axis of symmetry, and x-intercepts, and understanding how the coefficient 'a' affects the opening and width of the curve. Visualization of a quadratic function reinforces algebraic concepts and is key to interpreting real-world phenomena modeled by quadratic relationships.

Quadratic Functions

A quadratic function is a polynomial function of degree two that is typically written in the standard form y = ax² + bx + c. Understanding quadratic functions is essential as they are used to model parabolic shapes and can be analyzed for features such as the vertex, axis of symmetry, and roots. They are foundational in algebra and have applications across many areas of mathematics and science.

Systems of Equations

When a function must pass through several points, a system of equations can be constructed by substituting the coordinates of these points into the function’s equation. This method allows for the determination of unknown parameters by solving the simultaneous equations, an important skill not only in algebra but also in various applied mathematical problems.

Transcript

00:01 They give us those three points on this quadratic, and they want us to go ahead and figure out what the a, b, and c should be for this quadratic.

00:11 So the first thing they want us to do is set up the equation.

00:14 So let's see.

00:15 Well, if this is x and this is y, when we plug in 1 into that quadratic, we want to output of negative 2.

00:21 So we'd have negative 2 is equal to a times 1 squared plus b times 1 plus c.

00:28 So that would be negative 2 is equal to a plus b plus c.

00:34 Then we get a second equation by using the next point.

00:39 So we want an output of 0 when we plug in negative 1.

00:42 So it would be a times negative 1 squared plus b times negative 1 plus c.

00:48 And so that would give us a minus b plus c.

00:52 And then lastly, when x is 2, y should be 3.

00:55 So we have 3 is equal 2.

00:57 So a times 2 squared plus b times 2 plus c.

01:03 So that will give us 4a plus 2b plus c, and this is equal to 3.

01:08 So to set up the equation, it would be these three equations that i'm boxing right here.

01:16 Now for part b to actually solve, let's go ahead and set these up.

01:20 So i'm going to write them.

01:23 So a minus b plus c is equal to 0.

01:27 A plus b plus c is equal to negative 2, and 4a plus 2b plus c is equal to 3.

01:36 So it doesn't matter which order you write these in.

01:38 I just want to do it this way because when we start to add row 1 to row 2 and 3, we're just multiplying by 0.

01:44 So these won't change.

01:46 So it'll just save us a little bit of time.

01:48 All right, so let's see.

01:51 First, if we want to clear our row for the a's, so to clear the a and row 2 is going to be negative.

01:57 Row 1 plus row 2 and so that should give or actually let's write down the first row so it's still a minus b plus c is equal to 0 and now the a is cancel so we'd get just positive b add that to b so that give us plus 2b and then we get negative c add that to c that gives us 0 and then again anything multiply by 0 0 add that to negative 2 we just get negative 2 let's get rid of those zeros though so this should be the second equation now for our last equation we would want to multiply by four are one and then add that to row three to get that four a to cancel so the a is cancel negative one times negative four is four plus two would be six we have plus six b and then negative four c plus c would be negative three c and then again this won't change all right, so at this point, notice that we could just go ahead and solve for b and start back substituting.

03:12 So let's go ahead and do that.

03:16 So this implies that, so 2b is equal to negative 2, well, we would just divide by 2.

03:22 So b is negative 1...

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