Determin the $$ x $$ and $$ y $$ interapts for Quadratic equetion Ex-rcise (H/W) 1. Find the $$ x $$ and $$ y $$ interapt of the following. 4). $$ y=(x-3)(x+4) $$ f). $$ y=(x+3)^{2} $$ b). $$ y=x^{2}+5 x+6 $$ g). $$ y=x(x-5) $$ C). $$ 8 x^{2}-2 x-15-y=0 $$ 1). $$ y=(x+3)^{2}+2 $$ 1.). $$ x^{2}-2 x-8 $$ d. $$ y-6=(x-3)^{2} $$ e). $$ \frac{y}{-x+5}=(3 x+5) $$

Question

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To find the $$ x $$ and $$ y $$ intercepts for the given quadratic equations, we will follow these steps:

1. **Finding the $$ y $$-intercept**: Set $$ x = 0 $$ and solve for $$ y $$.
2. **Finding the $$ x $$-intercept**: Set $$ y = 0 $$ and solve for $$ x $$.

Let's go through each equation step by step.

### 1. $$ y = (x - 3)(x + 4) $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = (0 - 3)(0 + 4) = -3 \cdot 4 = -12
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = (x - 3)(x + 4)
  $$
  This gives $$ x - 3 = 0 $$ or $$ x + 4 = 0 $$, so $$ x = 3 $$ or $$ x = -4 $$.

### 2. $$ y = (x + 3)^{2} $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = (0 + 3)^{2} = 3^{2} = 9
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = (x + 3)^{2}
  $$
  This gives $$ x + 3 = 0 $$, so $$ x = -3 $$.

### 3. $$ y = x^{2} + 5x + 6 $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = 0^{2} + 5 \cdot 0 + 6 = 6
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = x^{2} + 5x + 6
  $$
  Factoring gives $$ (x + 2)(x + 3) = 0 $$, so $$ x = -2 $$ or $$ x = -3 $$.

### 4. $$ y = x(x - 5) $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = 0(0 - 5) = 0
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = x(x - 5)
  $$
  This gives $$ x = 0 $$ or $$ x = 5 $$.

### 5. $$ 8x^{2} - 2x - 15 - y = 0 $$

Rearranging gives $$ y = 8x^{2} - 2x - 15 $$.

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = 8(0)^{2} - 2(0) - 15 = -15
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = 8x^{2} - 2x - 15
  $$
  Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} $$:
  $$
  x = \frac{2 \pm \sqrt{(-2)^{2} - 4 \cdot 8 \cdot (-15)}}{2 \cdot 8}
  $$

### 6. $$ y = (x + 3)^{2} + 2 $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = (0 + 3)^{2} + 2 = 9 + 2 = 11
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = (x + 3)^{2} + 2
  $$
  This has no real solutions since the minimum value is 2.

### 7. $$ x^{2} - 2x - 8 = 0 $$

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = 0^{2} - 2(0) - 8 = -8
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = x^{2} - 2x - 8
  $$
  Using the quadratic formula:
  $$
  x = \frac{2 \pm \sqrt{(-2)^{2} - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}
  $$

### 8. $$ y - 6 = (x - 3)^{2} $$

Rearranging gives $$ y = (x - 3)^{2} + 6 $$.

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = (0 - 3)^{2} + 6 = 9 + 6 = 15
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = (x - 3)^{2} + 6
  $$
  This has no real solutions since the minimum value is 6.

### 9. $$ \frac{y}{-x + 5} = (3x + 5) $$

Rearranging gives $$ y = (3x + 5)(-x + 5) $$.

**Finding $$ y $$-intercept**:
- Set $$ x = 0 $$:
  $$
  y = (3(0) + 5)(-0 + 5) = 5 \cdot 5 = 25
  $$

**Finding $$ x $$-intercept**:
- Set $$ y = 0 $$:
  $$
  0 = (3x + 5)(-x + 5)
  $$
  This gives $$ 3x + 5 = 0 $$ or $$ -x + 5 = 0 $$, so $$ x = -\frac{5}{3} $$ or $$ x = 5 $$.

Now, let's summarize the intercepts for each equation:

1. $$ y = (x - 3)(x + 4) $$: $$ x $$-
Here are the $$ x $$ and $$ y $$ intercepts for each quadratic equation: 1. $$ y = (x - 3)(x + 4) $$: - $$ y $$-intercept: $$ (0, -12) $$ - $$ x $$-intercepts: $$ (3, 0) $$ and $$ (-4, 0) $$ 2. $$ y = (x + 3)^2 $$: - $$ y $$-intercept: $$ (0, 9) $$ - $$ x $$-intercept: $$ (-3, 0) $$ 3. $$ y = x^2 + 5x + 6 $$: - $$ y $$-intercept: $$ (0, 6) $$ - $$ x $$-intercepts: $$ (-2, 0) $$ and $$ (-3, 0) $$ 4. $$ y = x(x - 5) $$: - $$ y $$-intercept: $$ (0, 0) $$ - $$ x $$-intercepts: $$ (0, 0) $$ and $$ (5, 0) $$ 5. $$ 8x^2 - 2x - 15 - y = 0 $$: - $$ y $$-intercept: $$ (0, -15) $$ - $$ x $$-intercepts: $$ \frac{2 \pm \sqrt{4 + 480}}{16} = \frac{2 \pm \sqrt{484}}{16} = \frac{2 \pm 22}{16} $$, so $$ x = \frac{24}{16} = 1.5 $$ and $$ x = \frac{-20}{16} = -1.25 $$ 6. $$ y = (x + 3)^2 + 2 $$: - $$ y $$-intercept: $$ (0, 11) $$ - $$ x $$-intercepts: None (since the minimum $$ y $$-value is 2) 7. $$ x^2 - 2x - 8 = 0 $$: - $$ y $$-intercept: $$ (0, -8) $$ - $$ x $$-intercepts: $$ \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} $$, so $$ x = 4 $$ and $$ x = -2 $$ 8. $$ y - 6 = (x - 3)^2 $$: - $$ y $$-intercept: $$ (0, 15) $$ - $$ x $$-intercepts: $$ (5, 0) $$ and $$ (-\frac{5}{3}, 0) $$ 9. $$ \frac{y}{-x + 5} = 3x + 5 $$: - $$ y $$-intercept: $$ (0, 25) $$ - $$ x $$-intercepts: $$ (5, 0) $$ and $$ (-\frac{5}{3}, 0) $$

Determin the and interapts for Quadratic equetion Ex-rcise (H/W)

Find the and interapt of the following.
4).
f).
b).
g).
C).
1).
1.).
d.
e).

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Determin the and interapts for Quadratic equetion Ex-rcise (H/W) Find the and interapt of the following. 4). f). b). g). C). 1). 1.). d. e).