1. The problem asks to find the slope $m$ and intercept $b$ of the linear functions and graph them. 2. For each function: - $f(x) = 4x - 3$: $m=4$, $b=-3$ - $f(x) = \frac{2}{3}x + 2$: $m=\frac{2}{3}$, $b=2$ - $f(x) = -\frac{1}{3}x + 5$: $m=-\frac{1}{3}$, $b=5$ 3. The quadratic functions are: - $y = 1x^2 + 2x + 1$ - $y = x^2 - 4x + 4$ 4. To find the axis of symmetry of a parabola $y = ax^2 + bx + c$, use the formula: $$x = -\frac{b}{2a}$$ 5. For $y = x^2 - 4x + 4$, $a=1$, $b=-4$, so: $$x = -\frac{-4}{2(1)} = \frac{4}{2} = 2$$ 6. The vertex corresponds to this $x$ value; evaluate $y$ at $x=2$: $$y = (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0$$ 7. The vertex is at $(2,0)$ and the axis of symmetry is $x = 2$. 8. For table values of $y = x^2 - 4x + 4$ at $x = -3, -2, -1, 0, 1$: - At $x = -3$, $y= (-3)^2 - 4(-3) + 4 = 9 + 12 + 4 = 25$ - $x = -2$, $y=4+8+4=16$ - $x = -1$, $y=1+4+4=9$ - $x=0$, $y=0+0+4=4$ - $x=1$, $y=1 - 4 + 4=1$ 9. Graph shapes: - The linear graphs have slopes $4$, $\frac{2}{3}$, and $-\frac{1}{3}$ with intercepts $-3$, $2$, and $5$, respectively (lines). - The quadratic graph $y = x^2 - 4x +4$ is a parabola opening upwards with vertex at $(2,0)$ and axis $x=2$. 10. Summary of requested values: - $m$ and $b$ for linear functions: $(4,-3)$, $(\frac{2}{3},2)$, $(-\frac{1}{3},5)$ - Axis of symmetry for quadratic $y=x^2 - 4x + 4$: $x=2$ - Vertex of quadratic: $(2,0)$ Final answers: - Linear: $f(x) = mx + b$ with identified slopes and intercepts. - Quadratic axis: $x=2$ - Quadratic vertex: $(2,0)$