- Each ordered pair gives one equation after substitution into \(y=ax^2+bx+c\). - Be especially careful when squaring a negative x-value. - Notice what happens to the variable terms when the x-coordinate is zero.

1. From \((-1,4)\): \(a-b+c=4\). 2. From \((0,1)\): \(c=1\). 3. From \((2,7)\): \(4a+2b+c=7\).

\(a-b+c=4\) \(c=1\) \(4a+2b+c=7\)

- A point lies on a graph when its two coordinates satisfy the function rule. - Take special care when squaring a negative input. - Check all three points; one successful check is not enough.

1. At \(x=-1\), \(f(-1)=1-1+1=1\). 2. At \(x=0\), \(f(0)=0+0+1=1\). 3. At \(x=2\), \(f(2)=4+2+1=7\). 4. All three given points satisfy the equation.

Yes: \(f(-1)=1\), \(f(0)=1\), and \(f(2)=7\).

- Which given input makes both terms containing \(a\) and \(b\) disappear? - Substitute the now-known constant into each remaining point equation. - Keep the minus sign when using a negative x-coordinate.

1. Substituting \((0, 3)\) gives \(c=3\). 2. Substituting \((1, 4)\) gives \(a+b+3=4\), so \(a+b=1\). 3. Substituting \((-1, 6)\) gives \(a-b+3=6\), so \(a-b=3\).

\(c=3\); remaining equations: \(a+b=1\) and \(a-b=3\).

- Which outputs belong to neighboring input values? - Imagine the first changes as a new short list. Are its values equally spaced?

1. The successive output changes are \(2-1=1\), \(5-2=3\), and \(10-5=5\). 2. The differences between them are \(3-1=2\) and \(5-3=2\). 3. These second differences are equal for equally spaced inputs.

a) \(1,3,5\) b) \(2,2\); both differences are the same.

- Substitute both positive and negative inputs carefully. - Which terms in the two equations stay the same? - How could you compare the two equations without first finding all three coefficients?

1. The conditions give \(4a+2b+c=6\) and \(4a-2b+c=-2\). 2. Subtract the second equation from the first to cancel \(a\) and \(c\): \(4b=8\). 3. Therefore, \(b=2\).

\(b=2\)

- Start with the point whose input is zero. - Write one equation from each remaining point and look at their signs. - Can you combine the two equations to remove one unknown?

1. The point \((0, 2)\) gives \(c=2\). 2. The other points give \(a-b+2=6\) and \(a+b+2=4\), or \(a-b=4\) and \(a+b=2\). 3. Adding yields \(2a=6\), so \(a=3\), and then \(b=-1\). 4. Thus \(f(x)=3x^2-x+2\). Substitution gives outputs \(6\), \(2\), and \(4\) at the stated inputs.

\(f(x)=3x^2-x+2\)

- Compare the squared inputs \(0^2\), \(1^2\), and \(2^2\) with the linear inputs. - Which terms might cancel when you compare changes over equal input steps? - What remains from the quadratic term after making the second comparison?

1. Substitute the inputs: \(f(0)=c\), \(f(1)=a+b+c\), and \(f(2)=4a+2b+c\). 2. The first output change is \(f(1)-f(0)=a+b\). The second is \(f(2)-f(1)=3a+b\). 3. Subtract them: \((3a+b)-(a+b)=2a\). The \(b\) and \(c\) terms do not affect this second difference.

a) \(f(0)=c\), \(f(1)=a+b+c\), \(f(2)=4a+2b+c\) b) \((f(2)-f(1))-(f(1)-f(0))=(3a+b)-(a+b)=2a\)

- The output at zero directly gives one coefficient. - Compare consecutive changes in the output instead of solving all three equations at once. - What would the changes look like for a linear function? How are these values different?

1. From \(f(0)=2\), we get \(c=2\). 2. The outputs change by \(1\) and then \(5\); their difference is \(4\), so for unit-spaced inputs \(2a=4\) and \(a=2\). 3. From \(f(1)=3\), we have \(2+b+2=3\), giving \(b=-1\). 4. The rule is \(f(x)=2x^2-x+2\); it gives \(f(2)=8\) as required.

\(f(x)=2x^2-x+2\)

- Compare the x- and y-coordinates of the three marked points. - What does a parabola's symmetry tell you when two points have the same height? - How can an equation in vertex form make use of the coordinates you found?

1. Points \((-1, 2)\) and \((3, 2)\) have equal heights, so the symmetry axis is halfway between their x-coordinates: \(x=1\). 2. The third point is \((1, -2)\), on the axis, so it is the vertex. 3. Write \(f(x)=a(x-1)^2-2\). Using \((-1, 2)\), we get \(2=a(-2)^2-2\), so \(4a=4\) and \(a=1\). 4. Thus \(f(x)=(x-1)^2-2\); it passes through all three points.

\(f(x)=(x-1)^2-2\)

- What happens to a factor when its zero is substituted? - Two functions can have the same zeros but different heights. How could you account for that? - Which condition checks that your factorization gives the correct height?

1. The zeros give \(f(x)=a(x+3)(x-1)\). 2. The point \((0, -6)\) gives \(-6=a(3)(-1)=-3a\), so \(a=2\). 3. Therefore, \(f(x)=2(x+3)(x-1)\), which has the stated zeros and y-intercept.

\(f(x)=2(x+3)(x-1)\)

- One of the x-coordinates makes both variable terms disappear. - Simplify both remaining equations before trying to solve them together. - What combination of the equations would leave only one unknown?

1. The point \((0,-2)\) gives \(c=-2\). 2. The other two points give \(a+b-2=1\) and \(9a+3b-2=19\), so \(a+b=3\) and \(3a+b=7\). 3. Subtract the equations to get \(2a=4\). Thus \(a=2\), and then \(b=1\). 4. Therefore \(f(x)=2x^2+x-2\). Its outputs at \(0,1,3\) are \(-2,1,19\), as required.

\(f(x)=2x^2+x-2\); the point checks give \(f(0)=-2\), \(f(1)=1\), and \(f(3)=19\).

- What does a point on the \(y\)-axis tell you about \(c\)? - Substitute each point into \(f(x) = ax^2 + bx + c\). - How can you solve the resulting system of equations? - Check the final function using all three points.

1. Substitute \(A(0, 4)\): \(c = 4\). 2. Substitute \(B(1, 3)\): \(a + b + 4 = 3\), so \(a + b = -1\). 3. Substitute \(C(2, 6)\): \(4a + 2b + 4 = 6\), so \(2a + b = 1\). 4. Subtract \(a + b = -1\) from \(2a + b = 1\): \(a = 2\). Then \(2 + b = -1\), so \(b = -3\). 5. Therefore, \(f(x) = 2x^2 - 3x + 4\).

\(a = 2\), \(b = -3\), \(c = 4\); \(f(x) = 2x^2 - 3x + 4\)

- Substitute each ordered pair into the same standard-form template. - Look for two equations whose subtraction isolates one coefficient efficiently. - Keep the coefficient values visible before writing the final function.

1. Substitution gives \(a+b+c=1\), \(4a+2b+c=0\), and \(a-b+c=9\). 2. Subtract the third equation from the first: \(2b=-8\), so \(b=-4\). 3. Substitute \(b=-4\): \(a+c=5\) and \(4a+c=8\). 4. Subtract to get \(3a=3\), so \(a=1\), then \(c=4\). 5. Therefore, \(f(x)=x^2-4x+4\).

System: \(a+b+c=1\), \(4a+2b+c=0\), \(a-b+c=9\) \(a=1\), \(b=-4\), \(c=4\) \(f(x)=x^2-4x+4\)

- Every given point makes the quadratic equation true. What equations do you get from the three points? - Compare the equations from the points with x-coordinates \(-1\) and \(1\). - After finding one coefficient, how can you reduce the system to two unknowns?

1. Substitute the three points into \(f(x)=ax^2+bx+c\) to get \(a-b+c=9\), \(a+b+c=1\), and \(4a+2b+c=3\). 2. Subtract the first equation from the second: \(2b=-8\), so \(b=-4\). 3. Substitute \(b=-4\) into the first and third equations: \(a+c=5\) and \(4a+c=11\). 4. Subtract: \(3a=6\), so \(a=2\). Then \(c=3\). 5. The function is \(f(x)=2x^2-4x+3\). Substitution confirms all three given points.

\(a=2\), \(b=-4\), \(c=3\); \(f(x)=2x^2-4x+3\).

- What does the point \(A(0, 5)\) tell you immediately about \(c\)? - Substitute the known value of \(c\) into the other two point equations. - How do you test whether a point lies on a graph?

1. Substituting \(A(0, 5)\) gives \(c = 5\). 2. Substitute \(B(2, 1)\): \(4a + 2b + 5 = 1\), so \(2a + b = -2\). 3. Substitute \(C(4, 5)\): \(16a + 4b + 5 = 5\), so \(4a + b = 0\). 4. Subtract the equations: \(2a = 2\), so \(a = 1\). Then \(4 + b = 0\), so \(b = -4\). 5. Thus, \(f(x) = x^2 - 4x + 5\). 6. Evaluate \(f(5) = 5^2 - 4 \cdot 5 + 5 = 10\). Since this matches the point's \(y\)-coordinate, \(D(5, 10)\) lies on the graph.

\(f(x) = x^2 - 4x + 5\). Yes, \(D(5, 10)\) lies on the graph.

- Which point gives \(c\) immediately? - Use the other two points to create a system for \(a\) and \(b\). - After finding standard form, how can completing the square produce vertex form? - What information becomes visible in vertex form?

1. Substituting \(A(0, -1)\) gives \(c = -1\). 2. Substitute \(B(2, 5)\): \(4a + 2b - 1 = 5\), so \(4a + 2b = 6\). 3. Substitute \(C(-2, -3)\): \(4a - 2b - 1 = -3\), so \(4a - 2b = -2\). 4. Add the two equations: \(8a = 4\), so \(a = 0.5\). Then \(2 + 2b = 6\), so \(b = 2\). 5. The standard form is \(f(x) = 0.5x^2 + 2x - 1\). 6. Complete the square: \(f(x) = 0.5(x^2 + 4x) - 1 = 0.5((x + 2)^2 - 4) - 1 = 0.5(x + 2)^2 - 3\).

Standard form: \(f(x) = 0.5x^2 + 2x - 1\) Vertex form: \(f(x) = 0.5(x + 2)^2 - 3\)

- Write one equation for each of the three points. - Compare the equations from \(P\) and \(Q\). Which variable is easy to isolate? - After finding \(a\), \(b\), and \(c\), how can you rewrite the function to reveal the vertex?

1. Substitute the points: \(a - b + c = -5\), \(a + b + c = 3\), \(4a + 2b + c = 1\). 2. Subtract the first equation from the second: \(2b = 8\), so \(b = 4\). 3. Substitute \(b = 4\): the second equation gives \(a + c = -1\), and the third gives \(4a + c = -7\). 4. Subtract the equations: \(3a = -6\), so \(a = -2\). Then \(c = 1\). 5. The standard form is \(f(x) = -2x^2 + 4x + 1\). 6. Complete the square: \(f(x) = -2(x^2 - 2x) + 1 = -2((x - 1)^2 - 1) + 1 = -2(x - 1)^2 + 3\). The vertex is \((1, 3)\).

Standard form: \(f(x) = -2x^2 + 4x + 1\) Vertex form: \(f(x) = -2(x - 1)^2 + 3\) Vertex: \((1, 3)\)

- Use elimination to remove variables one at a time. - How can the equations from \(x = -1\) and \(x = 1\) simplify the system? - When completing the square with an odd linear coefficient, what number should you add and subtract?

1. Substitute the points: \(a - b + c = 1\), \(a + b + c = -1\), \(4a + 2b + c = 1\). 2. Subtract the second equation from the first: \(-2b = 2\), so \(b = -1\). 3. Substitute \(b = -1\). The second equation gives \(a + c = 0\), so \(c = -a\). The third gives \(4a - 2 + c = 1\). Thus, \(3a = 3\), so \(a = 1\) and \(c = -1\). 4. The standard form is \(f(x) = x^2 - x - 1\). 5. Complete the square: \(f(x) = x^2 - x + 0.25 - 0.25 - 1 = (x - 0.5)^2 - 1.25\).

Standard form: \(f(x) = x^2 - x - 1\) Vertex form: \(f(x) = (x - 0.5)^2 - 1.25\)

- What do you notice about the \(y\)-coordinates of \(A\) and \(C\)? - Where is the axis of symmetry between two points at the same height? - Which form is easiest when the vertex is known?

1. Points \(A\) and \(C\) have the same \(y\)-coordinate, so the axis of symmetry is halfway between their \(x\)-coordinates: \(x = \frac{1 + 5}{2} = 3\). 2. Since \(B(3, 3)\) lies on that axis, it is the vertex. Vertex form is therefore the most efficient choice: \(y = a(x - 3)^2 + 3\). 3. Substitute \(A(1, 7)\): \(7 = a(1 - 3)^2 + 3\), so \(4a = 4\) and \(a = 1\). 4. The function is \(y = (x - 3)^2 + 3\), or in standard form, \(y = x^2 - 6x + 12\).

Vertex form is the most efficient starting point. \(y = (x - 3)^2 + 3\), or \(y = x^2 - 6x + 12\) in standard form.

- When no zeros or vertex are given, standard form is a useful starting point. - Substitute each point to create three equations. - Eliminate one variable at a time from the system.

1. Use standard form \(y = ax^2 + bx + c\) and substitute the points: \(a - b + c = 11\), \(a + b + c = 3\), \(4a + 2b + c = 5\). 2. Subtract the first equation from the second: \(2b = -8\), so \(b = -4\). 3. Substitute \(b = -4\): the second equation gives \(a + c = 7\), and the third gives \(4a + c = 13\). 4. Subtract the equations: \(3a = 6\), so \(a = 2\). Then \(c = 5\). 5. Therefore, \(y = 2x^2 - 4x + 5\).

\(y = 2x^2 - 4x + 5\)

- What does a point with \(x = 0\) tell you about \(c\)? - Which coefficient determines a parabola's width and opening direction? - Do the points show a symmetry that helps locate the vertex?

a) Substituting \(P(0, 3)\) gives \(c = 3\). From \(Q(1, 2)\), \(a + b = -1\). From \(R(2, 3)\), \(2a + b = 0\). Subtracting gives \(a = 1\), then \(b = -2\). Thus, \(f(x) = x^2 - 2x + 3\). b) Since \(|a| = 1\), the parabola has the same width as \(y = x^2\). c) Rewrite \(f(x) = (x - 1)^2 + 2\), so the vertex is \((1, 2)\).

a) \(f(x) = x^2 - 2x + 3\) b) It has the same width as \(y = x^2\) because \(|a| = 1\). c) \((1, 2)\)

- Two points have \(y = 0\). What does that tell you about the graph? - How can the third point determine the leading coefficient once the zeros are used as factors? - How do you test whether a point lies on a graph? - How can completing the square produce vertex form?

a) Since \(K\) and \(L\) are x-intercepts, write \(f(x)=a(x+3)(x-1)\). Use \(M(2,10)\): \(10=a(2+3)(2-1)=5a\), so \(a=2\). Expanding gives \(f(x)=2(x+3)(x-1)=2x^2+4x-6\). b) Complete the square: \(f(x)=2(x^2+2x)-6=2((x+1)^2-1)-6=2(x+1)^2-8\). c) Evaluate \(f(-2)=2(-2)^2+4(-2)-6=-6\). Therefore, \(D(-2,-6)\) lies on the graph.

a) \(f(x) = 2x^2 + 4x - 6\) b) \(f(x) = 2(x + 1)^2 - 8\) c) Yes, \(D(-2, -6)\) lies on the graph.

- What do you notice about the \(y\)-coordinates of \(A\) and \(C\)? - Which point lies on the axis of symmetry? - Which form is easiest when the vertex is known?

1. Points \(A\) and \(C\) have the same \(y\)-coordinate and are equally far from \(x = 0\), so the axis of symmetry is \(x = 0\). 2. Since \(B(0, 5)\) lies on the axis, it is the vertex. Vertex form is therefore the most efficient starting point: \(y = ax^2 + 5\). 3. Substitute \(C(2, 1)\): \(1 = 4a + 5\), so \(a = -1\). 4. Thus, the equation is \(y = -x^2 + 5\).

Vertex form is the best starting point, and the equation is \(y = -x^2 + 5\).

- Which point gives the \(y\)-intercept directly? - Substitute the other two points to create a system in \(a\) and \(b\). - How can you eliminate one variable from the two equations?

1. Substitute \(P_1(0, 1)\): \(c = 1\). 2. Substitute \(P_2(2, 5)\): \(4a + 2b + 1 = 5\), so \(2a + b = 2\). 3. Substitute \(P_3(4, 5)\): \(16a + 4b + 1 = 5\), so \(4a + b = 1\). 4. Subtract the equations: \(2a = -1\), so \(a = -0.5\). 5. Substitute into \(2a + b = 2\): \(-1 + b = 2\), so \(b = 3\). 6. Therefore, \(f(x) = -0.5x^2 + 3x + 1\).

\(f(x) = -0.5x^2 + 3x + 1\)

- Compare the x-coordinates of the two points with equal y-values. - A point on the parabola that lies on the symmetry axis is especially useful when it is the lowest visible point implied by the symmetric pair. - Once the vertex is known, vertex form uses only one unknown scale factor.

1. Points \(P\) and \(R\) have equal y-values and x-coordinates equally spaced around \(x=2\), so the axis of symmetry is \(x=2\). 2. Point \(Q=(2,0)\) lies on that axis and on the parabola, so it is the vertex. 3. Write \(f(x)=a(x-2)^2\). Using \(P(0,4)\), \(4=4a\), so \(a=1\). 4. Expand: \(f(x)=(x-2)^2=x^2-4x+4\). Thus, \(a=1\), \(b=-4\), and \(c=4\).

Axis of symmetry: \(x=2\), because \(P\) and \(R\) have the same y-value at equal horizontal distances from \(x=2\). Therefore, \(Q(2,0)\) is the vertex. \(a=1\), \(b=-4\), \(c=4\); \(f(x)=x^2-4x+4\).

- What does a \(y\)-intercept of \(5\) tell you about \(c\)? - How can the two additional points create two equations in \(a\) and \(b\)? - Which system-solving method is most efficient here?

1. A \(y\)-intercept of \(5\) gives \(c = 5\). 2. Substitute \(D(-1, 12)\): \(a - b + 5 = 12\), so \(a - b = 7\). 3. Substitute \(E(2, 9)\): \(4a + 2b + 5 = 9\), so \(2a + b = 2\). 4. Add the equations: \(3a = 9\), so \(a = 3\). 5. Substitute into \(a - b = 7\): \(3 - b = 7\), so \(b = -4\). 6. Therefore, \(p(x) = 3x^2 - 4x + 5\).

\(p(x) = 3x^2 - 4x + 5\)

- How do you compare slopes to test collinearity? - What value can you read directly from a point on the \(y\)-axis? - How can you solve the resulting two equations in \(a\) and \(b\)?

1. Compute two slopes: \(m_{AB} = \frac{2 - 10}{0 - (-2)} = -4\) and \(m_{BC} = \frac{2 - 2}{2 - 0} = 0\). Since the slopes are different, the points are not collinear. 2. Substituting \(B(0, 2)\) gives \(c = 2\). 3. Substitute \(A(-2, 10)\): \(4a - 2b + 2 = 10\), so \(4a - 2b = 8\). 4. Substitute \(C(2, 2)\): \(4a + 2b + 2 = 2\), so \(4a + 2b = 0\). 5. Add the equations: \(8a = 8\), so \(a = 1\). Then \(4 + 2b = 0\), so \(b = -2\). 6. Therefore, \(f(x) = x^2 - 2x + 2\).

The points are not collinear. The quadratic function is \(f(x) = x^2 - 2x + 2\).

- Compare the slopes from \(P\) to \(Q\) and from \(Q\) to \(R\). - Substitute each point into \(f(x) = ax^2 + bx + c\). - How can subtracting consecutive equations eliminate \(c\)?

1. The slopes are \(m_{PQ} = \frac{3 - 1}{2 - 1} = 2\) and \(m_{QR} = \frac{7 - 3}{3 - 2} = 4\). Since the slopes are different, the points are not collinear. 2. Substitute the points: \(a + b + c = 1\), \(4a + 2b + c = 3\), \(9a + 3b + c = 7\). 3. Subtract consecutive equations: \(3a + b = 2\) and \(5a + b = 4\). 4. Subtract these equations: \(2a = 2\), so \(a = 1\). Then \(3 + b = 2\), so \(b = -1\). 5. Substitute into \(a + b + c = 1\): \(c = 1\). 6. Therefore, \(f(x) = x^2 - x + 1\).

The points are not collinear. The quadratic function is \(f(x) = x^2 - x + 1\).

- Compare the slopes between consecutive pairs of points. - What do you notice about the \(y\)-coordinates of \(K\) and \(M\)? - Can symmetry identify the vertex before you solve for the equation? - Fractional coefficients are valid results.

1. The slopes are \(m_{KL} = \frac{-2 - 1}{1 - (-2)} = -1\) and \(m_{LM} = \frac{1 - (-2)}{4 - 1} = 1\). Since the slopes are different, the points are not collinear. 2. Points \(K\) and \(M\) have the same \(y\)-coordinate and are equally spaced from \(x = 1\), so \(L(1, -2)\) is the vertex. 3. Use vertex form: \(f(x) = a(x - 1)^2 - 2\). 4. Substitute \(K(-2, 1)\): \(1 = a(-3)^2 - 2\), so \(9a = 3\) and \(a = \frac{1}{3}\). 5. Expand: \(f(x) = \frac{1}{3}(x - 1)^2 - 2 = \frac{1}{3}x^2 - \frac{2}{3}x - \frac{5}{3}\).

The points are not collinear. The quadratic function is \(f(x) = \frac{1}{3}x^2 - \frac{2}{3}x - \frac{5}{3}\).

- Substitute each point into standard form. - How can subtracting equations eliminate \(c\)? - Solve the resulting two-equation system before finding the final coefficient.

1. Substitute the points: \(4a - 2b + c = 10\), \(a + b + c = -2\), \(9a + 3b + c = 0\). 2. Subtract the second equation from the first: \(3a - 3b = 12\), so \(a - b = 4\). 3. Subtract the second equation from the third: \(8a + 2b = 2\), so \(4a + b = 1\). 4. Add the two reduced equations: \(5a = 5\), so \(a = 1\). Then \(1 - b = 4\), so \(b = -3\). 5. Substitute into \(a + b + c = -2\): \(c = 0\). 6. Therefore, \(f(x) = x^2 - 3x\).

\(f(x) = x^2 - 3x\)

- Use the point with \(x=0\) first; it determines one coefficient immediately. - Substitute that coefficient into the other two equations before eliminating a second variable. - Check the final function against all three points, not just the two used last.

1. Substitution gives \(a+b+c=5\), \(a-b+c=1\), and \(c=2\). 2. Using \(c=2\), the first two equations become \(a+b=3\) and \(a-b=-1\). 3. Add them: \(2a=2\), so \(a=1\). 4. Then \(b=2\), and \(c=2\). 5. Therefore, \(y=x^2+2x+2\).

System: \(a+b+c=5\), \(a-b+c=1\), \(c=2\) \(a=1\), \(b=2\), \(c=2\) \(y=x^2+2x+2\)

- Be careful with signs when substituting a negative \(x\)-value. - Which coefficients match in the equations from \(A\) and \(C\)? - Which elimination step removes two variables at once?

a) Substituting the points gives \(4a - 2b + c = 10\), \(a + b + c = 1\), and \(4a + 2b + c = 6\). b) Subtract the equation from \(A\) from the equation from \(C\): \(4b = -4\), so \(b = -1\). c) Substitute \(b = -1\) into the equations from \(B\) and \(C\): \(a + c = 2\) and \(4a + c = 8\). Subtract to get \(3a = 6\), so \(a = 2\), then \(c = 0\). Therefore, \(y = 2x^2 - x\).

a) \(4a - 2b + c = 10\), \(a + b + c = 1\), and \(4a + 2b + c = 6\) b) Subtracting the equations from \(A\) and \(C\) gives \(4b = -4\), so \(b = -1\). c) \(y = 2x^2 - x\)

- What does \(c\) represent on the graph? - Can you simplify either equation before combining them? - Check the final equation with both \(T\) and \(U\).

1. Since \(S\) has \(x = 0\), substituting it immediately gives \(c = 4\). 2. Substitute \(T(1, 3)\): \(a + b + 4 = 3\), so \(a + b = -1\). 3. Substitute \(U(-2, -6)\): \(4a - 2b + 4 = -6\), so \(2a - b = -5\). 4. Add the two equations: \(3a = -6\), so \(a = -2\). 5. Substitute into \(a + b = -1\): \(b = 1\). 6. Therefore, \(a = -2\), \(b = 1\), \(c = 4\), and \(y = -2x^2 + x + 4\).

Point \(S(0, 4)\) gives \(c = 4\) immediately. The coefficients are \(a = -2\), \(b = 1\), and \(c = 4\), so \(y = -2x^2 + x + 4\).

- Read each marker from the grid before setting up any equation. - The point on the y-axis determines the constant term immediately. - Use the remaining two points for the other coefficients, then check all three plotted coordinates.

1. The graph shows \((-2,7)\), \((0,1)\), and \((1,4)\). 2. The point \((0,1)\) gives \(c=1\). 3. The other points give \(4a-2b+1=7\) and \(a+b+1=4\), so \(2a-b=3\) and \(a+b=3\). 4. Solving gives \(a=2\) and \(b=1\), hence \(f(x)=2x^2+x+1\). 5. Check: \(f(-2)=7\), \(f(0)=1\), and \(f(1)=4\).

Points: \((-2,7)\), \((0,1)\), \((1,4)\) \(f(x)=2x^2+x+1\) Verification: \(f(-2)=7\), \(f(0)=1\), \(f(1)=4\)

- Substitute the entire negative x-coordinate into both \(x\) and \(x^2\) before simplifying. - Compare Luis's first equation term by term with \(a(-2)^2+b(-2)+c\). - After correcting the system, eliminate one coefficient by subtracting equations rather than relying on a y-axis shortcut.

1. Luis did not square the x-coordinate in the \(ax^2\) term. For \(x=-2\), the correct equation is \(4a-2b+c=9\). 2. The other equations are \(a+b+c=0\) and \(9a+3b+c=4\). 3. Eliminating \(c\) gives \(3a-3b=9\) and \(8a+2b=4\). 4. Solving gives \(a=1\), \(b=-2\), and \(c=1\), so \(y=x^2-2x+1\).

Luis forgot to square \(-2\) in the \(ax^2\) term. Correct system: \(4a-2b+c=9\), \(a+b+c=0\), \(9a+3b+c=4\) Quadratic: \(y=x^2-2x+1\)

- How do the revenue increases change as the price rises by the same amount each time? - What do equally spaced inputs tell you about the quadratic coefficient? - After finding a rule, does it reproduce all three stated revenue values?

1. The revenue changes by \(15\) and then \(5\) as the price increases by \(\$1\). The second difference is \(-10\), so \(2a=-10\) and \(a=-5\). 2. From \(R(4)=160\), \(-5(16)+4b+c=160\), or \(4b+c=240\). 3. From \(R(5)=175\), \(-5(25)+5b+c=175\), or \(5b+c=300\). Subtracting gives \(b=60\) and then \(c=0\). 4. The model is \(R(p)=-5p^2+60p\). It also gives \(R(6)=180\). At \(p=7\), \(R(7)=-5(49)+60(7)=175\).

\(R(p)=-5p^2+60p\); predicted daily revenue at \(\$7\) per bouquet: \(\$175\).

- Substitute the three points into \(g(x) = ax^2 + bx + c\). - The point \((0, 0)\) immediately determines \(c\). - Use \(x = -\frac{b}{2a}\) to find each vertex. - The horizontal range is the positive zero.

a) Let \(g(x) = ax^2 + bx + c\). Since \(P(0, 0)\) lies on the graph, \(c = 0\). Substituting \(Q(5, 4)\) and \(R(10, 6)\) gives \(25a + 5b = 4\) and \(100a + 10b = 6\). Solving yields \(a = -0.04\) and \(b = 1\), so \(g(x) = -0.04x^2 + x\). b) For Rocket A, the vertex occurs at \(x = -\frac{1}{2(-0.05)} = 10\), and \(f(10) = 5\), so its vertex is \((10, 5)\). For Rocket B, the vertex occurs at \(x = -\frac{1}{2(-0.04)} = 12.5\), and \(g(12.5) = 6.25\), so its vertex is \((12.5, 6.25)\). Rocket B reaches a greater height. c) Factoring gives \(f(x) = x(-0.05x + 1)\), so its positive zero is \(20\). Also, \(g(x) = x(-0.04x + 1)\), so its positive zero is \(25\). Rocket B travels farther.

a) \(g(x) = -0.04x^2 + x\) b) Rocket A: vertex \((10, 5)\); Rocket B: vertex \((12.5, 6.25)\). Rocket B goes higher. c) Rocket A: \(20\,\text{m}\); Rocket B: \(25\,\text{m}\). Rocket B travels farther.