Understand the move. Then solve the problem.

1. Isolate the quadratic and linear terms.

   $$f(x)=2(x^2-6x)+13$$

2. Make a perfect square while keeping the expression equal.

   $$f(x)=2[(x-3)^2-9]+13=2(x-3)^2-5$$

3. The square is smallest at x = 3. The parabola opens upwards.

   $$(h,k)=(3,-5),\qquad \min f=-5$$

Vertex (3, −5); axis x = 3; minimum −5 at x = 3.

Checks and common pitfalls: Expanding 2(x − 3)² − 5 recovers the original function. The factor 2 stretches the graph vertically without changing the x-coordinate of the vertex.

1. Identify the coefficients and compute the discriminant.

   $$\Delta=(-6)^2-4(1)k=36-4k$$

2. A repeated root requires a zero discriminant.

   $$36-4k=0\Rightarrow k=9$$

3. Substitute the parameter and factor the expression.

   $$y=x^2-6x+9=(x-3)^2$$

k = 9; the graph touches the x-axis at (3, 0).

Checks and common pitfalls: For k < 9 there are two intersections; for k > 9 there are none. Changing k moves the graph vertically while its axis stays at x = 3.

1. Express the other side in terms of x.

   $$\ell=14-x$$

2. Multiply the side lengths and complete the square.

   $$A(x)=x(14-x)=49-(x-7)^2$$

3. The squared term is nonnegative, and x = 7 is in the allowed domain.

   $$A(x)\le49,\qquad A(7)=49$$

The greatest area is 49 m², attained by a 7 m × 7 m square.

Checks and common pitfalls: The downward-opening area graph has vertex (7, 49). Checking the domain matters: a negative or zero side length would not describe the display.

1. Rewrite the bracket to show h.

   $$x+2=x-(-2)$$

2. Read the vertex. The negative leading coefficient changes the opening direction.

   $$(h,k)=(-2,7)$$

B: the vertex is (−2, 7).

Checks and common pitfalls: The graph opens downwards and its maximum is 7. The coefficient −3 is not a coordinate of the vertex.

1. Evaluate the discriminant.

   $$\Delta=4^2-4(1)(8)=-16<0$$

2. Check the conclusion using the vertex form.

   $$x^2+4x+8=(x+2)^2+4\ge4$$

D: there are no real roots.

Checks and common pitfalls: The whole graph lies above the x-axis. A negative discriminant does not mean the function is negative.

1. Factor the quadratic.

   $$2x^2-7x+3=(2x-1)(x-3)$$

2. A product is zero when at least one factor is zero.

   $$2x-1=0\ \text{or}\ x-3=0\Rightarrow x=\tfrac12\ \text{or}\ 3$$

C: the roots are 1/2 and 3.

Checks and common pitfalls: Their sum is 7/2 and product is 3/2, agreeing with −b/a and c/a. Both values also give zero on substitution.

1. Move the vertex with the graph.

   $$(1,2)\longmapsto(1-3,2-4)=(-2,-2)$$

2. Keep the opening and scale; use the new vertex.

   $$y=(x-(-2))^2-2=(x+2)^2-2$$

A: y = (x + 2)² − 2.

Checks and common pitfalls: A left shift replaces x by x + 3 in the original expression. It does not replace x by x − 3.

1. Complete the square.

   $$f(x)=3[(x+2)^2-4]+5=3(x+2)^2-7$$

2. Set the nonnegative square to zero.

   $$f(-2)=-7=\min f$$

The minimum is −7, reached at x = −2.

Checks and common pitfalls: The required output is the vertical coordinate of the vertex. Entering −2 gives where the minimum occurs, not the minimum value.

1. Use the relationships between roots and coefficients.

   $$\alpha+\beta=5,\qquad \alpha\beta=2$$

2. Express the required quantity using that sum and product.

   $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=25-4=21$$

The sum of the squares is 21.

Checks and common pitfalls: Squaring the sum introduces 2αβ. Leaving that term in would give 25, which is not the sum of the squares.

1. Factor the difference of squares.

   $$(x-4)(x+4)=0$$

2. Select the positive root.

   $$x=4$$

The positive root is 4.

Checks and common pitfalls: The full solution set is {−4,4}; the word positive restricts the requested answer.

1. Bound the squared term.

   $$-2(x-1)^2\le0$$

2. Equality is attained at x=1.

   $$\max y=8$$

Maximum 8 at x=1.

Checks and common pitfalls: The sign of the leading coefficient determines whether the vertex is a maximum or minimum.

1. Substitute the coefficients.

   $$\Delta=2^2-4(1)(5)$$

2. A negative value excludes real roots.

   $$\Delta=-16<0$$

Discriminant −16; no real roots.

Checks and common pitfalls: A negative discriminant does not mean there are no complex roots.

1. The zero parameter gives a linear equation.

   $$m=0\Rightarrow x=2$$

2. Classify the quadratic using the discriminant.

   $$m\ne0:\quad \Delta=9+24m$$

3. Keep the linear exception distinct.

   $$m<-3/8:0;\quad m=-3/8:1;\quad m>-3/8,\ m\ne0:2$$

m=0 has one linear root; m=−3/8 a repeated root x=4; the other counts follow Δ.

Checks and common pitfalls: The condition a≠0 is essential; using the quadratic formula at m=0 would divide by zero.

1. Use the degree-to-radian conversion.

   $$150^\circ=150\cdot\frac{\pi}{180}=\frac{5\pi}{6}\ \text{rad}$$

2. Reflect the unit-circle point across the vertical axis.

   $$\sin150^\circ=\sin(180^\circ-30^\circ)=\sin30^\circ$$

3. Read the exact vertical coordinate.

   $$\sin\frac{5\pi}{6}=\frac12$$

150° = 5π/6 radians, and its sine is 1/2.

Checks and common pitfalls: Reflection across the vertical axis preserves the sine value and changes the sign of cosine. A negative sine here would place the point below the horizontal axis.

1. Amplitude is the magnitude of the vertical multiplier.

   $$|A|=|-2|=2,\qquad D=1$$

2. Solve for a change in x that produces a full cycle.

   $$3T=2\pi\Rightarrow T=\frac{2\pi}{3}$$

3. Add the vertical shift after scaling the range.

   $$-1\le\cos(3x)\le1\Rightarrow -1\le y\le3$$

Amplitude 2; period 2π/3; midline y = 1; range [−1, 3].

Checks and common pitfalls: The negative multiplier reflects the cosine curve in its midline. At x = 0 the curve starts at its minimum −1; the amplitude is still positive.

1. Read the maximum height from the midline and amplitude.

   $$h_{\max}=4+3=7$$

2. Solve the peak condition for time.

   $$\frac{\pi}{6}(t-2)=\frac\pi2+2\pi n\Rightarrow t=5+12n,\quad n\in\mathbb Z$$

3. Check the observation interval: the preceding peak is before t = 0.

   $$t=-7,\ 5,\ 17,\ldots\Rightarrow t_{\mathrm{first}}=5$$

The first maximum is at t = 5 hours, when the height is 7 m.

Checks and common pitfalls: The model has period 12 hours and a right shift of 2 hours. The shift marks an upward midline crossing, not the first maximum. This is an invented model, not a measurement record.

1. Identify the reference angle.

   $$\frac{2\pi}{3}=\pi-\frac\pi3$$

2. Reflect the horizontal coordinate on the unit circle.

   $$\cos\left(\pi-\frac\pi3\right)=-\cos\frac\pi3=-\frac12$$

B: −1/2.

Checks and common pitfalls: The positive √3/2 is the sine at this angle. Distinguish the horizontal coordinate (cosine) from the vertical coordinate (sine).

1. Set the change in the argument to one full turn.

   $$2(x+T)-2x=2\pi$$

2. Solve for the period in x.

   $$2T=2\pi\Rightarrow T=\pi$$

C: the least positive period is π radians.

Checks and common pitfalls: In a width of 2π along the x-axis this curve completes two cycles. Its amplitude is 3, independently of its period.

1. Match the angle to x − h.

   $$h=\frac\pi4$$

2. Read the vertical translation and check a reference point.

   $$(0,0)\longmapsto\left(\frac\pi4,1\right)$$

D: shift right by π/4 and up by 1.

Checks and common pitfalls: At x = π/4 the angle inside sine is zero and the output is 1. This checks the direction of both translations.

1. Recover the vertical parameters.

   $$D=\frac{8+2}{2}=5,\qquad A=\frac{8-2}{2}=3$$

2. Recover the horizontal multiplier from the period.

   $$B=\frac{2\pi}{4\pi}=\frac12$$

3. A positive cosine coefficient gives the required initial peak.

   $$y(0)=3\cos0+5=8$$

A: y = 3 cos(x/2) + 5.

Checks and common pitfalls: B confuses amplitude with the full vertical span; C uses the wrong period; D starts at the minimum. Each stated feature helps rule out a different error.

1. Write the full-cycle condition in degrees.

   $$4T=360$$

2. Divide by the horizontal multiplier.

   $$T=\frac{360}{4}=90$$

The least positive period is 90°.

Checks and common pitfalls: The amplitude 2 and downward translation 3 do not change the period. The radian value π/2 represents the same angular interval but is not the requested numerical unit.

1. Set the argument to the peak angles of sine.

   $$2x-\frac\pi3=\frac\pi2+2\pi n$$

2. Solve the linear equation for x.

   $$x=\frac{5\pi}{12}+\pi n,\quad n\in\mathbb Z$$

3. The preceding solution is negative; divide the first positive solution by π.

   $$c=\frac5{12}\approx0.416666667$$

c = 5/12 ≈ 0.416666667, so the first positive peak is at x = 5π/12.

Checks and common pitfalls: The horizontal shift is π/6, and one quarter of the period is π/4. Their sum is 5π/12. For a decimal answer, use at least six decimal places.

1. Apply the conversion.

   $$60^\circ=\frac{60}{180}\pi$$

2. Read the coefficient of π.

   $$c=\frac13$$

c=1/3.

Checks and common pitfalls: The requested number is the coefficient, not the radian value itself.

1. Use opposite divided by hypotenuse.

   $$\sin30^\circ=\frac12$$

2. Enter a fraction or decimal.

   $$\frac12=0.5$$

1/2.

Checks and common pitfalls: The angle is in degrees; 30 radians is a different angle.

1. Account for the horizontal scale.

   $$T=\frac{2\pi}{3}$$

2. Read the coefficient.

   $$c=\frac23$$

c=2/3.

Checks and common pitfalls: Frequency 3 divides the period; it does not multiply it.

1. One valid counterexample disproves an all-real claim.

   $$\sin(3\pi/2)=-1\ne1=\sin(\pi/2)$$

2. Apply the compound-angle identity.

   $$\sin(x+\pi)=\sin x\cos\pi+\cos x\sin\pi=-\sin x$$

Counterexample x=π/2; correct identity sin(x+π)=−sin x.

Checks and common pitfalls: A half-turn changes the sign; the full period of sine is 2π.