Math  /  Geometry

Question[y+2=12(x+52]\left[y+2=\frac{1}{2}\left(x+5^{2}\right] \quad\right. SKETCH A) VERTEX: B) ×1NT\times 1 N T \qquad c) YINT: =5+z1=\frac{5+z}{-1} p) (UR lDOWN (s)RCHE ONE) ). max/min:ym a x / m i n: y : \qquad ق) Axis of Symm T TRY: \qquad 5) Domnon: \qquad \qquad H). RAWGE: \qquad -

Studdy Solution

STEP 1

What is this asking? We're asked to analyze a parabola, find some key points, and sketch its graph! Watch out! Don't forget that the vertex form of a parabola helps us see the vertex directly, but it's *h,kh, k*, not *x,yx, y*!

STEP 2

1. Rewrite in vertex form
2. Find the vertex
3. Find the x-intercept(s)
4. Find the y-intercept
5. Determine the direction
6. Find the maximum/minimum y-value
7. Find the axis of symmetry
8. Find the domain
9. Find the range

STEP 3

We want our equation to look like y=a(xh)2+ky = a(x - h)^2 + k, so let's **subtract** 2\bf{2} from both sides of the given equation y+2=12(x+5)2y + 2 = \frac{1}{2}(x + 5)^2: y=12(x+5)22 y = \frac{1}{2}(x + 5)^2 - 2

STEP 4

To really nail the vertex form, let's rewrite (x+5)(x+5) as (x(5))(x - (-5)): y=12(x(5))22 y = \frac{1}{2}(x - (-5))^2 - 2 Now, it perfectly matches y=a(xh)2+ky = a(x - h)^2 + k!

STEP 5

From our vertex form equation y=12(x(5))22y = \frac{1}{2}(x - (-5))^2 - 2, we see that h=5\bf{h = -5} and k=2\bf{k = -2}.

STEP 6

So, our **vertex** is at (5,2)\bf{(-5, -2)}!

STEP 7

To find the x-intercept(s), we set y=0y = 0: 0=12(x+5)22 0 = \frac{1}{2}(x + 5)^2 - 2

STEP 8

**Add** 2\bf{2} to both sides: 2=12(x+5)2 2 = \frac{1}{2}(x + 5)^2 **Multiply** both sides by 2\bf{2}: 4=(x+5)2 4 = (x + 5)^2 Take the **square root** of both sides: ±2=x+5 \pm 2 = x + 5 So, x+5=2x + 5 = 2 or x+5=2x + 5 = -2.
This gives us x=3 \bf{x = -3} and x=7 \bf{x = -7}.

STEP 9

Our **x-intercepts** are (3,0)\bf{(-3, 0)} and (7,0)\bf{(-7, 0)}.

STEP 10

Let's set x=0x = 0 in our equation y=12(0+5)22y = \frac{1}{2}(0 + 5)^2 - 2: y=12(5)22 y = \frac{1}{2}(5)^2 - 2

STEP 11

y=2522 y = \frac{25}{2} - 2 y=25242 y = \frac{25}{2} - \frac{4}{2} y=212 y = \frac{21}{2}

STEP 12

Our **y-intercept** is (0,212)\bf{(0, \frac{21}{2})} or (0,10.5)\bf{(0, 10.5)}.

STEP 13

Since a=12a = \frac{1}{2} is **positive**, the parabola opens **upwards**.

STEP 14

Since the parabola opens upwards, the vertex represents the **minimum** point.

STEP 15

The **minimum** y\bf{y}-value is 2\bf{-2}.

STEP 16

The axis of symmetry is a vertical line passing through the vertex, so its equation is x=hx = h.

STEP 17

In our case, the **axis of symmetry** is x=5\bf{x = -5}.

STEP 18

The domain of a parabola is all real numbers.

STEP 19

Our **domain** is (,)\bf{(-\infty, \infty)}.

STEP 20

Since our parabola opens upwards, the range starts from the minimum y-value and goes to infinity.

STEP 21

Our **range** is [2,)\bf{[-2, \infty)}.

STEP 22

Vertex: (5,2)(-5, -2) X-intercepts: (3,0)(-3, 0) and (7,0)(-7, 0) Y-intercept: (0,212)(0, \frac{21}{2}) Direction: Upwards Minimum y: 2-2 Axis of Symmetry: x=5x = -5 Domain: (,)(-\infty, \infty) Range: [2,)[-2, \infty)

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