7. Determine the $$ y $$-intercept, zeros, equation of the axis of symmetry, and vertex of each quadratic relation. Then sketch its graph. a) $$ y=(x-5)(x+5) $$ b) $$ y=-(x-6)(x-2) $$ c) $$ y=2(x-1)(x+3) $$ d) $$ y=-0.5(x+4)^{2} $$

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**a) $$y = (x-5)(x+5)$$**  
- **Zeros:** Set each factor equal to zero:  
  $$
  x-5=0 \Rightarrow x=5,\quad x+5=0 \Rightarrow x=-5.
  $$
- **Y-intercept:** Substitute $$x=0$$ into the equation:  
  $$
  y=(0-5)(0+5)=(-5)(5)=-25.
  $$
- **Axis of symmetry:** The axis is the average of the zeros:  
  $$
  x=\frac{5+(-5)}{2}=0.
  $$
- **Vertex:** The vertex lies on the axis. Substitute $$x=0$$ into the equation:  
  $$
  y=(0-5)(0+5)=-25 \quad \Longrightarrow \quad (0,-25).
  $$

---

**b) $$y = -(x-6)(x-2)$$**  
- **Zeros:**  
  $$
  x-6=0 \Rightarrow x=6,\quad x-2=0 \Rightarrow x=2.
  $$
- **Y-intercept:** Substitute $$x=0$$ into the equation:  
  $$
  y=-((0-6)(0-2))=-((-6)(-2))=-12.
  $$
- **Axis of symmetry:**  
  $$
  x=\frac{6+2}{2}=\frac{8}{2}=4.
  $$
- **Vertex:** Substitute $$x=4$$ into the equation:  
  $$
  y=-((4-6)(4-2))=-((-2)(2))=-(-4)=4.
  $$
  The vertex is $$(4,4)$$.

---

**c) $$y = 2(x-1)(x+3)$$**  
- **Zeros:**  
  $$
  x-1=0 \Rightarrow x=1,\quad x+3=0 \Rightarrow x=-3.
  $$
- **Y-intercept:** Substitute $$x=0$$ into the equation:  
  $$
  y=2(0-1)(0+3)=2(-1)(3)=-6.
  $$
- **Axis of symmetry:**  
  $$
  x=\frac{1+(-3)}{2}=\frac{-2}{2}=-1.
  $$
- **Vertex:** Substitute $$x=-1$$ into the equation:  
  $$
  y=2((-1)-1)((-1)+3)=2(-2)(2)= -8.
  $$
  The vertex is $$(-1,-8)$$.

---

**d) $$y = -0.5(x+4)^2$$**  
- **Vertex:** The equation is in vertex form $$y=a(x-h)^2+k$$ with $$a=-0.5$$, $$h=-4$$, and $$k=0$$. Thus, the vertex is  
  $$
  (-4,0).
  $$
- **Y-intercept:** Substitute $$x=0$$ into the equation:  
  $$
  y=-0.5(0+4)^2=-0.5(16)=-8.
  $$
- **Zeros:** Set $$y=0$$:  
  $$
  -0.5(x+4)^2=0 \quad \Longrightarrow \quad (x+4)^2=0 \quad \Longrightarrow \quad x=-4.
  $$
  This is a double root.
- **Axis of symmetry:**  
  $$
  x=-4.
  $$

---

*Graph Sketch Notes:*  
- **For (a):** The parabola $$y=x^2-25$$ opens upward, crosses the $$x$$-axis at $$-5$$ and $$5$$, and has a vertex $$(0,-25)$$.  
- **For (b):** The parabola opens downward (due to the negative coefficient), has zeros at $$x=2$$ and $$x=6$$, a vertex at $$(4,4)$$, and a $$y$$-intercept at $$-12$$.  
- **For (c):** Opens upward (multiplied by $$2$$), with zeros at $$-3$$ and $$1$$, vertex at $$(-1,-8)$$, and $$y$$-intercept at $$-6$$.  
- **For (d):** A downward-opening parabola with its vertex at $$(-4,0)$$, a double zero at $$x=-4$$, and a $$y$$-intercept at $$-8$$.
**a) $$y = (x-5)(x+5)$$** - **Zeros:** $$x = 5$$ and $$x = -5$$ - **Y-intercept:** $$(0, -25)$$ - **Axis of symmetry:** $$x = 0$$ - **Vertex:** $$(0, -25)$$ **b) $$y = -(x-6)(x-2)$$** - **Zeros:** $$x = 6$$ and $$x = 2$$ - **Y-intercept:** $$(0, -12)$$ - **Axis of symmetry:** $$x = 4$$ - **Vertex:** $$(4, 4)$$ **c) $$y = 2(x-1)(x+3)$$** - **Zeros:** $$x = 1$$ and $$x = -3$$ - **Y-intercept:** $$(0, -6)$$ - **Axis of symmetry:** $$x = -1$$ - **Vertex:** $$(-1, -8)$$ **d) $$y = -0.5(x+4)^2$$** - **Vertex:** $$(-4, 0)$$ - **Y-intercept:** $$(0, -8)$$ - **Zero:** $$x = -4$$ (double root) - **Axis of symmetry:** $$x = -4$$

Determine the -intercept, zeros, equation of the
axis of symmetry, and vertex of each quadratic
relation. Then sketch its graph.
a)
b)
c)
d)

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