Question

1. for each function, state the locator point (h,k), then sketch the graph.

a. $f(x) = 2|x + 1| - 3$ b. $f(x) = (x - 3)^2 + 4$ c. $f(x) = -x^3 - 3$

friday, jan. 30

1. factor the expression.

a. $3x^2 + 11x + 10$ b. $6x^3 - 31x^2 + 5x$ c. $6ab^2 + 15ab - 21a$ d. $y^2 + 5y - 24$

1. write the equation of the graph below.

a. image of a circle b. image of a line

Explanation:

Problem 7

Step1: Identify vertex form for a.

The absolute value vertex form is $f(x)=a|x-h|+k$. Rewrite $f(x)=2|x+1|-3$ as $f(x)=2|x-(-1)|+(-3)$. So $(h,k)=(-1,-3)$.

Step2: Identify vertex form for b.

The quadratic vertex form is $f(x)=a(x-h)^2+k$. For $f(x)=(x-3)^2+4$, $(h,k)=(3,4)$.

Step3: Identify vertex form for c.

Rewrite $f(x)=-x^2-3$ as $f(x)=-(x-0)^2+(-3)$. So $(h,k)=(0,-3)$.
(Graph sketch notes: a. V-shaped graph opening up, vertex at (-1,-3); b. Parabola opening up, vertex at (3,4); c. Parabola opening down, vertex at (0,-3))

Step1: Factor quadratic a.

Find two numbers: $5$ and $6$ (product $30$, sum $11$).
$3x^2+11x+10=3x^2+6x+5x+10=3x(x+2)+5(x+2)$

Step2: Factor cubic b.

First factor out $x$, then factor quadratic.
$6x^3-31x^2+5x=x(6x^2-31x+5)=x(6x^2-30x-x+5)=x[6x(x-5)-1(x-5)]$

Step3: Factor expression c.

First factor out GCF $3a$, then simplify.
$6ab^2+15ab-21a=3a(2b^2+5b-7)=3a(2b^2+7b-2b-7)=3a[b(2b+7)-1(2b+7)]$

Step4: Factor quadratic d.

Find two numbers: $8$ and $-3$ (product $-24$, sum $5$).
$y^2+5y-24=y^2+8y-3y-24=y(y+8)-3(y+8)$

Step1: Find circle equation a.

Circle standard form: $(x-h)^2+(y-k)^2=r^2$. Center $(h,k)=(1,4)$, radius $r=4$ (from center to x-axis intercept).
Substitute values: $(x-1)^2+(y-4)^2=4^2$

Step2: Find line equation b.

Use two-point form: $(x_1,y_1)=(-3,-3)$, $(x_2,y_2)=(6,0)$.
Slope $m=\frac{0-(-3)}{6-(-3)}=\frac{3}{9}=\frac{1}{3}$. Use point-slope $y-y_1=m(x-x_1)$: $y-0=\frac{1}{3}(x-6)$

Answer:

a. Locator point: $(-1, -3)$
b. Locator point: $(3, 4)$
c. Locator point: $(0, -3)$

---

Problem 8