Question

topic: recognizing quadratic equations
identify whether or not each equation represents a quadratic function. explain how you knew it was a quadratic.

1. $x^{2}+13x - 4 = 0$

quadratic or no?
justification:

1. $3x^{2}+x = 3x^{2}-2$

quadratic or no?
justification:

1. $x(4x - 5)=0$

quadratic or no?
justification:

1. $(2x - 7)+6x = 10$

quadratic or no?
justification:

1. $2^{x}+6 = 0$

quadratic or no?
justification:

1. $32 = 4x^{2}$

quadratic or no?
justification:
set
topic: changing from standard form of a quadratic to vertex form.
change the form of each equation to vertex form: $y = a(x - h)^{2}+k$. state the vertex and graph the parabola. show at least 3 accurate points on each side of the line of symmetry.

1. $y = x^{2}-4x + 1$

vertex:

1. $y = x^{2}+2x + 5$

vertex:

Explanation:

Step1: Recall quadratic - function form

A quadratic function is of the form $ax^{2}+bx + c=0$ where $a
eq0$.

Step2: Analyze equation 1

For $x^{2}+13x - 4=0$, $a = 1$, $b = 13$, $c=-4$. Since $a = 1
eq0$, it is a quadratic.
Quadratic or no? Quadratic
Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 1
eq0$.

Step3: Analyze equation 2

Simplify $3x^{2}+x=3x^{2}-2$. Subtract $3x^{2}$ from both sides: $x=-2$, which is a linear equation.
Quadratic or no? No
Justification: After simplification, the $x^{2}$ - terms cancel out, leaving a linear equation.

Step4: Analyze equation 3

Expand $x(4x - 5)=0$ to get $4x^{2}-5x=0$. Here $a = 4$, $b=-5$, $c = 0$. Since $a = 4
eq0$, it is a quadratic.
Quadratic or no? Quadratic
Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 4
eq0$.

Step5: Analyze equation 4

Simplify $(2x - 7)+6x=10$ to $8x-7 = 10$ or $8x=17$, which is linear.
Quadratic or no? No
Justification: There is no $x^{2}$ - term after simplification.

Step6: Analyze equation 5

The equation $2^{x}+6 = 0$ is an exponential equation, not a quadratic as the variable $x$ is in the exponent.
Quadratic or no? No
Justification: It is an exponential equation, not of the form $ax^{2}+bx + c = 0$.

Step7: Analyze equation 6

Rewrite $32 = 4x^{2}$ as $4x^{2}-32=0$ with $a = 4$, $b = 0$, $c=-32$. Since $a = 4
eq0$, it is a quadratic.
Quadratic or no? Quadratic
Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 4
eq0$.

Step8: Convert $y=x^{2}-4x + 1$ to vertex - form

Complete the square:
\[

$$\begin{align*} y&=x^{2}-4x + 1\\ y&=(x^{2}-4x+4)-4 + 1\\ y&=(x - 2)^{2}-3 \end{align*}$$

\]
The vertex is $(h,k)=(2,-3)$.
vertex: $(2,-3)$

Step9: Convert $y=x^{2}+2x + 5$ to vertex - form

Complete the square:
\[

$$\begin{align*} y&=x^{2}+2x + 5\\ y&=(x^{2}+2x + 1)-1+5\\ y&=(x + 1)^{2}+4 \end{align*}$$

\]
The vertex is $(h,k)=(-1,4)$.
vertex: $(-1,4)$

Answer:

1. Quadratic; Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 1

eq0$.

1. No; Justification: After simplification, the $x^{2}$ - terms cancel out, leaving a linear equation.
2. Quadratic; Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 4

eq0$.

1. No; Justification: There is no $x^{2}$ - term after simplification.
2. No; Justification: It is an exponential equation, not of the form $ax^{2}+bx + c = 0$.
3. Quadratic; Justification: It is in the form $ax^{2}+bx + c = 0$ with $a = 4

eq0$.

1. vertex: $(2,-3)$
2. vertex: $(-1,4)$