1. **Problem 1: Sketch the line with y-intercept -4 and slope 5/3.** - The line equation in slope-intercept form is $y = mx + b$ where $m=\frac{5}{3}$ and $b=-4$. - So, the equation is $$y = \frac{5}{3}x - 4.$$ - This is a straight line starting at $(0,-4)$ and rising 5 units for every 3 units run. 2. **Problem 2: Sketch the parabola opening upward with vertex (-3, 2), passing through (-5,5) and (1,5).** - Vertex form of a parabola: $y = a(x - h)^2 + k$ with vertex $(h,k) = (-3,2)$. - Use point $(-5,5)$ to find $a$: $$5 = a(-5 + 3)^2 + 2 \implies 5 = a( -2 )^2 + 2 = 4a + 2$$ $$4a = 3 \implies a = \frac{3}{4}$$ - Equation: $$y = \frac{3}{4}(x + 3)^2 + 2$$ - Double-check with point $(1,5)$: $$y = \frac{3}{4}(1 + 3)^2 + 2 = \frac{3}{4} \times 16 + 2 = 12 + 2 = 14$$ - Since $14 \neq 5$, check the other point $ (1,5)$ instead by solving for $a$ using both points: Set system: $$5 = a(-5+3)^2 + 2 = 4a + 2$$ $$5 = a(1+3)^2 + 2 = 16a + 2$$ - First: $4a + 2 = 5 \Rightarrow 4a=3 \Rightarrow a=\frac{3}{4}$ - Second: $16a + 2 = 5 \Rightarrow 16a=3 \Rightarrow a=\frac{3}{16}$ - Contradiction: So, both points can't satisfy same $a$ with given vertex and form. - Use standard form $y = ax^2 + bx + c$ with vertex formula $h = -\frac{b}{2a} = -3$ and $k=2$. - From vertex: $b = -2ah$; $c = k - ah^2$. - Plug points $(-5,5)$ and $(1,5)$: $$5 = a(-5)^2 + b(-5) + c = 25a -5b + c$$ $$5 = a(1)^2 + b(1) + c = a + b + c$$ - Also, from vertex $b = 6a$ (since $b = -2a(-3) = 6a$), and $c = 2 - 9a$. - Substitute $b$ and $c$: $$5 = 25a - 5(6a) + 2 - 9a = 25a -30a + 2 -9a = (-14a) + 2$$ $$5 - 2 = -14a \Rightarrow 3 = -14a \Rightarrow a = -\frac{3}{14}$$ - Then, $b = 6a = 6 \times -\frac{3}{14} = -\frac{18}{14} = -\frac{9}{7}$ - And $c = 2 - 9a = 2 - 9 \times -\frac{3}{14} = 2 + \frac{27}{14} = \frac{28}{14} + \frac{27}{14} = \frac{55}{14}$. - Final equation: $$y = -\frac{3}{14}x^2 - \frac{9}{7}x + \frac{55}{14}$$ 3. **Problem 3: Sketch $y = -2x + 4$.** - A straight line with slope $-2$ and y-intercept $4$. 4. **Problem 4: Sketch parabola $f(x) = x^2 + 4x + 4$.** - Factorize: $$f(x) = (x + 2)^2$$ - Vertex at $(-2,0)$ and parabola opens upward. 5. **Problem 5: Point (-1, 2) with slope -4.** - Equation of tangent line through point $(x_1,y_1) = (-1,2)$ with slope $m = -4$: $$y - 2 = -4(x + 1) \Rightarrow y = -4x - 4 + 2 = -4x - 2$$ **Summary:** 1. Line: $y = \frac{5}{3}x - 4$ 2. Parabola: $y = -\frac{3}{14}x^2 - \frac{9}{7}x + \frac{55}{14}$ 3. Line: $y = -2x + 4$ 4. Parabola: $y = (x + 2)^2$ 5. Line through $(-1,2)$ with slope $-4$: $y = -4x - 2$