Question

f(x) = -x² + 7x

1. reinforce write the symbolic representation in factored, vertex, and general forms for each quadratic function graphed here.

a.
factored form:
standard or general form:
vertex form:
b.
factored form:
standard or general form:
vertex form:

Explanation:

To solve for the factored, standard, and vertex forms of the quadratic functions, we analyze each graph:

Left Graph (First Quadratic)

Step 1: Identify x - intercepts (roots)

The parabola crosses the x - axis at \( x=-2 \) and \( x = 1 \). For a quadratic function, if the roots are \( r_1\) and \( r_2\), the factored form is \( y=a(x - r_1)(x - r_2) \). Here, \( r_1=-2 \) and \( r_2 = 1 \), so factored form is \( y=a(x + 2)(x - 1) \).

To find \( a \), we use the y - intercept. The parabola passes through \( (0,-2) \) (since when \( x = 0 \), \( y=-2 \)):
Substitute \( x = 0 \), \( y=-2 \) into \( y=a(x + 2)(x - 1) \):
\( -2=a(0 + 2)(0 - 1)\Rightarrow -2=a(2)(-1)\Rightarrow -2=-2a\Rightarrow a = 1 \).

• Factored form: \( \boldsymbol{y=(x + 2)(x - 1)} \)

Step 2: Expand to standard form (\( y=ax^2+bx + c \))

Expand \( (x + 2)(x - 1) \):
\( y=x^2 - x+2x - 2=x^2+x - 2 \).

• Standard form: \( \boldsymbol{y=x^2+x - 2} \)

Step 3: Convert to vertex form (\( y=a(x - h)^2+k \), where \( (h,k) \) is the vertex)

Complete the square for \( y=x^2+x - 2 \):
\( x^2+x=x^2+x+\frac{1}{4}-\frac{1}{4}=(x+\frac{1}{2})^2-\frac{1}{4} \)
So, \( y=(x+\frac{1}{2})^2-\frac{1}{4}-2=(x+\frac{1}{2})^2-\frac{9}{4} \).

• Vertex form: \( \boldsymbol{y=(x+\frac{1}{2})^2-\frac{9}{4}} \)

Right Graph (Second Quadratic)

Step 1: Identify the vertex and a point

The vertex is at \( (0,-4) \), so vertex form is \( y=a(x - 0)^2-4=ax^2-4 \).

To find \( a \), use the x - intercept (e.g., \( (2,0) \)):
Substitute \( x = 2 \), \( y = 0 \) into \( y=ax^2-4 \):
\( 0=a(2)^2-4\Rightarrow 4a=4\Rightarrow a = 1 \).

• Vertex form: \( \boldsymbol{y=\frac{1}{4}x^2-4} \) (Wait, correction: If we re - check, if the vertex is \( (0,-4) \) and it passes through \( (4,0) \), let's redo: Let the vertex form be \( y=a(x - 0)^2-4=ax^2-4 \). Substitute \( (4,0) \): \( 0=a(4)^2-4\Rightarrow 16a=4\Rightarrow a=\frac{1}{4} \). So vertex form is \( \boldsymbol{y=\frac{1}{4}x^2-4} \))

Step 2: Expand to standard form

\( y=\frac{1}{4}x^2-4 \) (already in standard form since there is no \( bx \) term).

• Standard form: \( \boldsymbol{y=\frac{1}{4}x^2-4} \)

Step 3: Find factored form

Set \( y = 0 \): \( \frac{1}{4}x^2-4=0\Rightarrow x^2=16\Rightarrow x=\pm4 \). So roots are \( x=-4 \) and \( x = 4 \).
Factored form: \( y=\frac{1}{4}(x + 4)(x - 4) \) (since \( a=\frac{1}{4} \) and roots are \( \pm4 \)).

• Factored form: \( \boldsymbol{y=\frac{1}{4}(x + 4)(x - 4)} \)

Final Summary:

Form	Left Graph	Right Graph
Standard	\( y=x^2+x - 2 \)	\( y=\frac{1}{4}x^2-4 \)
Vertex	\( y=(x+\frac{1}{2})^2-\frac{9}{4} \)	\( y=\frac{1}{4}x^2-4 \) (or \( y=\frac{1}{4}(x - 0)^2-4 \))

(Note: The handwritten answers in the image had minor errors; the above is the corrected derivation.)

Answer:

To solve for the factored, standard, and vertex forms of the quadratic functions, we analyze each graph:

Left Graph (First Quadratic)

Step 1: Identify x - intercepts (roots)

The parabola crosses the x - axis at \( x=-2 \) and \( x = 1 \). For a quadratic function, if the roots are \( r_1\) and \( r_2\), the factored form is \( y=a(x - r_1)(x - r_2) \). Here, \( r_1=-2 \) and \( r_2 = 1 \), so factored form is \( y=a(x + 2)(x - 1) \).

To find \( a \), we use the y - intercept. The parabola passes through \( (0,-2) \) (since when \( x = 0 \), \( y=-2 \)):
Substitute \( x = 0 \), \( y=-2 \) into \( y=a(x + 2)(x - 1) \):
\( -2=a(0 + 2)(0 - 1)\Rightarrow -2=a(2)(-1)\Rightarrow -2=-2a\Rightarrow a = 1 \).

• Factored form: \( \boldsymbol{y=(x + 2)(x - 1)} \)

Step 2: Expand to standard form (\( y=ax^2+bx + c \))

Expand \( (x + 2)(x - 1) \):
\( y=x^2 - x+2x - 2=x^2+x - 2 \).

• Standard form: \( \boldsymbol{y=x^2+x - 2} \)

Step 3: Convert to vertex form (\( y=a(x - h)^2+k \), where \( (h,k) \) is the vertex)

Complete the square for \( y=x^2+x - 2 \):
\( x^2+x=x^2+x+\frac{1}{4}-\frac{1}{4}=(x+\frac{1}{2})^2-\frac{1}{4} \)
So, \( y=(x+\frac{1}{2})^2-\frac{1}{4}-2=(x+\frac{1}{2})^2-\frac{9}{4} \).

• Vertex form: \( \boldsymbol{y=(x+\frac{1}{2})^2-\frac{9}{4}} \)

Right Graph (Second Quadratic)

Step 1: Identify the vertex and a point

The vertex is at \( (0,-4) \), so vertex form is \( y=a(x - 0)^2-4=ax^2-4 \).

To find \( a \), use the x - intercept (e.g., \( (2,0) \)):
Substitute \( x = 2 \), \( y = 0 \) into \( y=ax^2-4 \):
\( 0=a(2)^2-4\Rightarrow 4a=4\Rightarrow a = 1 \).

• Vertex form: \( \boldsymbol{y=\frac{1}{4}x^2-4} \) (Wait, correction: If we re - check, if the vertex is \( (0,-4) \) and it passes through \( (4,0) \), let's redo: Let the vertex form be \( y=a(x - 0)^2-4=ax^2-4 \). Substitute \( (4,0) \): \( 0=a(4)^2-4\Rightarrow 16a=4\Rightarrow a=\frac{1}{4} \). So vertex form is \( \boldsymbol{y=\frac{1}{4}x^2-4} \))

Step 2: Expand to standard form

\( y=\frac{1}{4}x^2-4 \) (already in standard form since there is no \( bx \) term).

• Standard form: \( \boldsymbol{y=\frac{1}{4}x^2-4} \)

Step 3: Find factored form

Set \( y = 0 \): \( \frac{1}{4}x^2-4=0\Rightarrow x^2=16\Rightarrow x=\pm4 \). So roots are \( x=-4 \) and \( x = 4 \).
Factored form: \( y=\frac{1}{4}(x + 4)(x - 4) \) (since \( a=\frac{1}{4} \) and roots are \( \pm4 \)).

• Factored form: \( \boldsymbol{y=\frac{1}{4}(x + 4)(x - 4)} \)

Final Summary:

Form	Left Graph	Right Graph
Standard	\( y=x^2+x - 2 \)	\( y=\frac{1}{4}x^2-4 \)
Vertex	\( y=(x+\frac{1}{2})^2-\frac{9}{4} \)	\( y=\frac{1}{4}x^2-4 \) (or \( y=\frac{1}{4}(x - 0)^2-4 \))

(Note: The handwritten answers in the image had minor errors; the above is the corrected derivation.)