Free Math Worksheets

- Expand the product, then combine the \(x^2\)-terms, x-terms, and constants. - Substitute each input into the simplified expression. - A point lies on the graph when its coordinates satisfy the equation.

1. Expand and combine like terms: \(f(x)=2x^2-6x+x-3-2x^2+4x=-x-3\). 2. \(f(-5)=5-3=2\), \(f(-3)=3-3=0\), \(f(0)=-3\), and \(f(2)=-2-3=-5\). 3. \(f(12)=-12-3=-15\), which matches the y-coordinate of \(P\). Therefore, \(P\) lies on the graph.

a) \(f(x)=-x-3\) b) \(f(-5)=2\), \(f(-3)=0\), \(f(0)=-3\), \(f(2)=-5\) c) Yes.

- Expand the square and distribute the negative sign carefully. - A zero occurs where \(f(x)=0\). - For part c, set the function equal to \(12\) and solve.

1. Expand and combine like terms: \(f(x)=x^2-8x+16-x^2+10x-15=2x+1\). Because the simplified expression has the form \(mx+b\), \(f\) is linear. 2. Solve \(2x+1=0\). Then \(x=-\frac{1}{2}\). 3. Solve \(12=2x+1\). Then \(11=2x\), so \(x=\frac{11}{2}=5.5\).

a) \(f(x)=2x+1\) b) \(x=-\frac{1}{2}\) c) \(x=\frac{11}{2}\)

- Use the distributive property to remove parentheses. - Combine x-terms separately from constant terms. - Steepness depends on the absolute value of the slope. - To evaluate a function, substitute the given input for \(x\).

1. Simplify \(f\): \(f(x) = 2x - 6 + 4 = 2x - 2\). Thus \(m = 2\) and \(b = -2\). 2. Simplify \(g\): \(g(x) = -\frac{1}{2}x + 5 - 2 = -\frac{1}{2}x + 3\). Thus \(m = -\frac{1}{2}\) and \(b = 3\). 3. Steepness depends on the absolute value of the slope. Since \(|2| > \left|-\frac{1}{2}\right|\), the graph of \(f\) is steeper. 4. \(f(10) = 2 \cdot 10 - 2 = 18\).

a) \(f(x) = 2x - 2\); \(g(x) = -\frac{1}{2}x + 3\) b) \(f\): \(m = 2\), \(b = -2\); \(g\): \(m = -\frac{1}{2}\), \(b = 3\) c) \(f\) is steeper because \(|2| > \left|-\frac{1}{2}\right|\). d) \(f(10) = 18\)