Question

1. give the equation of a quadratic function, in vertex form, for a parabola that has a vertex at (3,16) and has a y - intercept at - 2. the general form of a quadratic in vertex form is: $y = a(x - h)^2 + k$

solve by substitution or elimination

1. $3x + 2y = 24$

$x - y = - 2$
solve by graphing

1. $y = \frac{3}{4}x + 2$

$2y = - x + 14$

Explanation:

(Problem 10):

Step1: Plug vertex into vertex form

The vertex is $(3,16)$, so $h=3$, $k=16$. Substitute into $y=a(x-h)^2+k$:
$y=a(x-3)^2+16$

Step2: Use y-intercept to find $a$

Y-intercept is $(0,-2)$. Substitute $x=0$, $y=-2$:
$-2=a(0-3)^2+16$
$-2=9a+16$
$9a=-2-16=-18$
$a=\frac{-18}{9}=-2$

Step3: Write final vertex form

Substitute $a=-2$, $h=3$, $k=16$ back:
$y=-2(x-3)^2+16$

(Problem 11 - Substitution Method):

Step1: Isolate $x$ from second equation

From $x-y=-2$, rearrange to get:
$x=y-2$

Step2: Substitute $x$ into first equation

Replace $x$ in $3x+2y=24$:
$3(y-2)+2y=24$
$3y-6+2y=24$
$5y=24+6=30$
$y=\frac{30}{5}=6$

Step3: Find $x$ using $y=6$

Substitute $y=6$ into $x=y-2$:
$x=6-2=4$

(Problem 12 - Graphing[LLM SSE On Failure]

Answer:

(Problem 12 - Graphing[LLM SSE On Failure]