for each of the equations below, complete the following:\n\n• find the x - and y - intercepts.\n\n• find the…

a. $y = x^{2}+8x + 12$

Explanation:

Step1: Find x - intercepts

Set $y = 0$, then $x^{2}+8x + 12=0$. Factor: $(x + 2)(x+6)=0$. So $x=-2$ or $x=-6$.

Step2: Find y - intercept

Set $x = 0$, then $y=0^{2}+8\times0 + 12=12$.

Step3: Find vertex

For $y=ax^{2}+bx + c$ ($a = 1$, $b = 8$, $c = 12$), the x - coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{8}{2\times1}=-4$. Substitute $x=-4$ into the equation: $y=(-4)^{2}+8\times(-4)+12=16 - 32+12=-4$. So the vertex is $(-4,-4)$.

Step4: Write in graphing form

Complete the square: $y=x^{2}+8x + 12=(x + 4)^{2}-16 + 12=(x + 4)^{2}-4$.

Answer:

• x - intercepts: $x=-2,x=-6$
• y - intercept: $y = 12$
• vertex: $(-4,-4)$
• graphing form: $y=(x + 4)^{2}-4$

b. $y=(x - 4)(x + 2)$

Explanation:

Step1: Find x - intercepts

Set $y = 0$, then $(x - 4)(x + 2)=0$. So $x = 4$ or $x=-2$.

Step2: Find y - intercept

Set $x=0$, then $y=(0 - 4)(0 + 2)=-8$.

Step3: Find vertex

First expand to $y=x^{2}-2x - 8$. For $y=ax^{2}+bx + c$ ($a = 1$, $b=-2$, $c=-8$), the x - coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{-2}{2\times1}=1$. Substitute $x = 1$ into the equation: $y=1^{2}-2\times1-8=1-2 - 8=-9$. So the vertex is $(1,-9)$.

Step4: Write in graphing form

Complete the square: $y=x^{2}-2x - 8=(x - 1)^{2}-1-8=(x - 1)^{2}-9$.

Answer:

• x - intercepts: $x = 4,x=-2$
• y - intercept: $y=-8$
• vertex: $(1,-9)$
• graphing form: $y=(x - 1)^{2}-9$

c. $y=x^{2}-6x - 9$

Explanation:

Step1: Find x - intercepts

Set $y = 0$, then use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $x^{2}-6x - 9=0$ ($a = 1$, $b=-6$, $c=-9$). $x=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-9)}}{2\times1}=\frac{6\pm\sqrt{36 + 36}}{2}=\frac{6\pm\sqrt{72}}{2}=\frac{6\pm6\sqrt{2}}{2}=3\pm3\sqrt{2}$.

Step2: Find y - intercept

Set $x = 0$, then $y=0^{2}-6\times0-9=-9$.

Step3: Find vertex

The x - coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{-6}{2\times1}=3$. Substitute $x = 3$ into the equation: $y=3^{2}-6\times3-9=9-18 - 9=-18$. So the vertex is $(3,-18)$.

Step4: Write in graphing form

Complete the square: $y=x^{2}-6x - 9=(x - 3)^{2}-9-9=(x - 3)^{2}-18$.

Answer:

• x - intercepts: $x=3 + 3\sqrt{2},x=3-3\sqrt{2}$
• y - intercept: $y=-9$
• vertex: $(3,-18)$
• graphing form: $y=(x - 3)^{2}-18$

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

Explanation:

Step1: Find x - intercepts

Set $y = 0$, then use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $x^{2}+5x + 1=0$ ($a = 1$, $b = 5$, $c = 1$). $x=\frac{-5\pm\sqrt{5^{2}-4\times1\times1}}{2\times1}=\frac{-5\pm\sqrt{25 - 4}}{2}=\frac{-5\pm\sqrt{21}}{2}$.

Step2: Find y - intercept

Set $x = 0$, then $y=0^{2}+5\times0+1=1$.

Step3: Find vertex

The x - coordinate of the vertex is $x=-\frac{b}{2a}=-\frac{5}{2\times1}=-\frac{5}{2}$. Substitute $x=-\frac{5}{2}$ into the equation: $y=\left(-\frac{5}{2}\right)^{2}+5\times\left(-\frac{5}{2}\right)+1=\frac{25}{4}-\frac{25}{2}+1=\frac{25 - 50 + 4}{4}=-\frac{21}{4}$. So the vertex is $\left(-\frac{5}{2},-\frac{21}{4}\right)$.

Step4: Write in graphing form

Complete the square: $y=x^{2}+5x + 1=\left(x+\frac{5}{2}\right)^{2}-\frac{25}{4}+1=\left(x+\frac{5}{2}\right)^{2}-\frac{21}{4}$.

Answer:

• x - intercepts: $x=\frac{-5+\sqrt{21}}{2},x=\frac{-5-\sqrt{21}}{2}$
• y - intercept: $y = 1$
• vertex: $\left(-\frac{5}{2},-\frac{21}{4}\right)$
• graphing form: $y=\left(x+\frac{5}{2}\right)^{2}-\frac{21}{4}$