Unit 7 Introduction To Quadratic Functions — Unit Plan

1. The missing value in the second row is 600. The missing values in the last row are 10 and 400.
2. Sample response: The points would form a stretched out upside-down U shape, or an arch.
3. Sample response: As one quantity increases, the other quantity first increases and then decreases, instead of always increasing or always decreasing.

Pattern B shows a quadratic relationship. Sample explanations:

• The number of dots is the step number squared plus 1, or $n^2+1$.
• (Students might answer by elimination.) Pattern A grows linearly. It is growing by 2 each time. Pattern C is growing by a factor of 3 each time, so it grows exponentially.
• (Students might recall the pattern they saw in the lesson but not yet internalize the squared input.) Pattern B is growing by 1, then by 3, then by 5, which is the pattern we saw in other examples of quadratic relationships in the lesson.

1. $y = n^2 + n$ (or equivalent), because there is an $n$-by-$n$ array of small squares, which gives $n^2$, and then there is a row of $n$ squares in the lower right.
2. The relationship is a quadratic function because the number of squares could be expressed as $n^2+n$, which is a quadratic expression.

Tyler is not correct. For small values of $x$, $5x^2$ is greater than $2^x$, but $2^x$ is eventually greater because it always doubles when $x$ increases by 1. When $x = 9$, $5(9)^2 = 405$, and $2^{9} = 512$, so $2^x$ is larger.

1. 842, or $906 - 64$ (or equivalent) 
2. $906-16t^2$

1. The graph intersects the vertical axis at 5. 
2. It indicates that the initial velocity of the rocket was 100 feet per second upward.
3. This indicates the effect of gravity pulling the rocket back toward Earth.
4. At about 6.3 seconds.

1. $5, because it would generate about $3,700 in revenue. Charging $17 would bring in about $2,500.
2. Based on the graph, a price of $10 would bring in about $5,000.
3. $0 \leq p \leq 20$, because the price has to be non-negative, and the company would earn negative revenue if the price were more than $20.

1. Diagram should show partial products of $3x(x)$, $6x$, $x$, and 2, which add up to $3x^2 + 7x+2$.
2. No. $(x+4)(x+4) = x^2 + 4x + 4x + 16 = x^2 + 8x + 16$. (Students may use a diagram or distributive property to reason.)

1. $2x^2 + 7x + 5$. Sample reasoning: The diagram shows that $(2x+5)(x+1) = 2x \boldcdot x + 2x \boldcdot 1 + 5 \boldcdot x + 5\boldcdot 1$. Adding this together gives $2x^2 + 7x + 5$ in standard form.

   	$2x$	$5$
   $x$	$2x^2$	$5x$
   $1$	$2x$	$5$

2. $x^2 - 4$. Sample reasoning: $(x+2)(x-2) = x^2 + (\text-2x) +2x + (\text-4) = x^2 - 4$. The diagram also shows that $(x+2)(x-2) = x^2-4$.

   	$x$	$2$
   $x$	$x^2$	$2x$
   $\text-2$	$\text-2x$	$\text-4$

Sample responses:

The $y$-intercept of the graph can be seen when the equation is in standard form. The $y$-coordinate is the number without the variable. In the equation $y=x^2+6x+8$, it is the 8, so the $y$-intercept is $(0,8)$.

The numbers in the factors of an equation in factored form give a clue about where the $x$-intercepts are when the equation is graphed. In the equation $y=(x+2)(x+4)$, the $x$-intercepts are $(\text-2,0)$ and $(\text-4,0)$.

1. $(2,0)$ and $(\text-4,0)$
2. $x = \text-1$, $y = (\text-1-2)(\text-1+4) = \text-9$
3. $y$-intercept is $(0,\text-8)$, because $(0-2)(0+4) = \text-8$.
4. The graph opens upward with vertex at $(\text-1, \text-9)$ and $x$-intercepts at $(2,0)$ and $(\text-4,0)$.

   

1. Sample response:
   • Graph A represents $h$ because the $y$-intercept is $(0,4)$, and there are no $x$-intercepts because this function always takes positive values.
   • Graph B represents $f$ because the $y$-intercept is $(0,\text-4)$, and the graph opens upward.
   • Graph C represents $g$ because the $y$-intercept is $(0,1)$, and the graph opens downward.
2. $y=1 - x^2$. The negative coefficient of $x^2$ makes the graph open downward.

1. The $x$-intercepts are at $x = 0$ and $x = 4$, with the vertex halfway between the intercepts, at $x = 2$.
2. Sample reasoning: $x^2+5x$ is equivalent to $x(x+5)$, so the $x$-intercepts are at $x=0$ and $x=\text-5$. The squared term has a positive coefficient so the graph opens upward. Adding a positive linear term to $x^2$ shifts the graph down and to the left.

1. 1 and $\text- \frac18$. Sample reasoning:
   • From the factored form, we can tell that the graph intersects the horizontal axis at $t=1$ and $t=\text-\frac18$. The zeros of the function are those same values.
   • (If students graph the function) The graph shows $t$-intercepts at $(\text- \frac18,0)$ and $(1,0)$.
2. The zeros mean the time in seconds when the ball hits the ground. Only one of them is meaningful. The negative zero doesn’t mean anything here.
3. From a height of 2 feet. Sample reasoning:
   • The height when $t=0$ is $(\text-16(0)-2)(0-1) = (\text-2)(\text-1) = 2$.
   • (If students graph the function) The graph shows a $y$-intercept of $(0,2)$.

1. The vertex of $f$ is at $(\text-3,\text-1)$.

2. The graph opens upward. The squared term has a positive coefficient (1).

1. $(3,2)$
2. Sample response: $(0,11)$ and $(6,11)$. When $x$ is 0, the value of $y$ is $(\text-3)^2+2$, or 11. The other point is 3 units to the right of the vertex with the same $y$-coordinate.

3. 

1. A graph the same shape as $y = x^2$ but shifted 6 units to the right and 10 units up. Its vertex is at $(6,10)$.
2. $y = \text-(x+2)^2-10$