Quadratic functions (Mathematics I)

From y=(x−2)²−4+1=(x−2)²−3 the vertex is (2, −3). For (x−2)² the x-coordinate of the vertex is +2, with the sign reversed. The option (2, 1) comes from forgetting to subtract 4, and (2, 5) from reading −4 as +4.

y=(x+3)²−9+5=(x+3)²−4. Since (x+3)²=(x−(−3))², the x-coordinate of the vertex is −3, so the vertex is (−3, −4). The option (3, −4) misreads the sign, and (−3, 5) drops the −9.

2(x²−4x)+3=2{(x−2)²−4}+3=2(x−2)²−8+3=2(x−2)²−5. The value −1 comes from forgetting to multiply the −4 by the coefficient 2 when bringing it out (−4+3=−1). Expanding to check that you return to 2x²−8x+3 is worthwhile.

From y=(x+1)²−1 the axis is the line x=−1. The axis sits at half the coefficient of x with the sign changed: half of 2 is 1, and changing the sign gives −1. The option x=1 is a sign error, and x=−2 comes from forgetting to halve.

−(x²−4x)−1=−{(x−2)²−4}−1=−(x−2)²+4−1=−(x−2)²+3, so the vertex is (2, 3). The option (2, −5) comes from mishandling the sign of −4 and computing −4−1. Take care with signs when removing a bracket that was factored out with −1.

A shift of p in the x-direction replaces x by x−p in the equation, so a shift of +3 gives (x−3)². The y-direction shift is −2, so −2 is added at the end. The key point of translation is that the sign in the equation is opposite to the shift, and (x+3)² comes from forgetting that reversal.

A shift of −1 in the x-direction replaces x by x−(−1)=x+1, giving 2(x+1)², and the y-direction shift of +4 adds +4 at the end. Checking with the vertex, (0,0) moves to (−1,4), which matches the vertex (−1,4) of y=2(x+1)²+4.

Reflection in the x-axis replaces y by −y, giving y=−(x²−2x+3)=−x²+2x−3, so every term changes sign. The option y=x²+2x+3 is the result of reflection in the y-axis (x replaced by −x), which is easy to mix up.

Reflection in the origin replaces x by −x and y by −y. From −y=(−x)²−4(−x)+5=x²+4x+5 we get y=−x²−4x−5. The option y=−x²+4x−5 applies only the x-axis reflection and y=x²+4x+5 only the y-axis reflection, each forgetting one substitution.

Write the vertex form y=a(x−1)²+2 and substitute (0, 4): a+2=4, so a=2. The curve y=2(x+1)²+2 also passes through (0, 4), but its vertex is (−1, 2), which fails the condition. And y=(x−1)²+2 gives 3 at x=0, so it does not pass through the point.

From y=(x−1)²+2 the parabola opens upward with vertex (1, 2), so the minimum is 2, attained at x=1. With no restriction on the domain, the minimum is exactly the y-coordinate of the vertex. The value 3 is what you get by substituting x=0, which is not the minimum.

From y=−(x−2)²+4 the parabola opens downward with vertex (2, 4), so the maximum is 4, attained at x=2. A downward parabola attains its maximum at the vertex and has no minimum. A sign error gives −4, but substituting x=2 gives −4+8=4, which settles it.

The equation is already in vertex form with vertex (3, 1). Since a=2>0 the parabola opens upward, so the minimum is 1, attained at x=3. Do not misread the vertex coordinates (3, 1): x=3 is where it happens and 1 is the minimum. The value 19 is the value at x=0.

The axis x=1 lies inside the domain, so the maximum is attained at the endpoint farther from the axis, x=3: f(3)=(3−1)²+2=4+2=6. The value 2 is the minimum, at the vertex. The value 18 comes from a sign error computing (3+1)².

The axis x=2 lies to the right of the domain 0≦x≦1, so the function decreases throughout the interval. The minimum is at the right endpoint x=1: f(1)=1−4+1=−2. The value −3 is the y-coordinate of the vertex, but x=2 is outside the domain, so it cannot be attained. Answering with the vertex while ignoring the domain is the typical mistake.

y=−(x−1)²+3. The parabola opens downward and the axis x=1 lies inside the domain, so the maximum is at the vertex, f(1)=3. The minimum is at the endpoint farther from the axis, x=3: f(3)=−9+6+2=−1. The endpoint value f(0)=2 is larger, because x=3 is farther from the axis.

The axis x=3 lies to the left of the domain 4≦x≦6, so the function increases throughout the interval. The minimum is at the left endpoint x=4: f(4)=16−24+10=2. The value 1 is the vertex value, but x=3 is outside the domain. The maximum is 10, at the right endpoint x=6.

Since a>2, the axis x=a lies to the right of the domain, so the function decreases throughout the interval and the minimum is at the right endpoint x=2: f(2)=4−4a. The value −a² is the vertex value, but x=a is outside the domain and cannot be attained. For example, with a=3 we get f(2)=−8, matching 4−4×3=−8.

The endpoint farther from the axis x=1 is x=3, so the maximum is f(3)=9−6+c=3+c. From 3+c=7 we get c=4. The option c=7 comes from wrongly taking the maximum to be f(0)=c, and c=8 from setting the minimum f(1)=c−1 equal to 7. With c=4 you can check that f(3)=7.

Putting t=x²−2x gives t=(x−1)²−1≧−1. Then y=t²−4t+1=(t−2)²−3, and t=2 lies within t≧−1, so the minimum is −3, at t=2, that is when x²−2x=2. At the endpoint t=−1 of the range we get y=6, which is not the minimum. After a substitution, never forget to check the range of t.

From (x−2)(x−3)=0 we get x=2, 3. The two numbers with product +6 and sum −5 are −2 and −3, so the factors are (x−2)(x−3) and the solutions come back with the opposite signs, 2 and 3. Answering x=−2, −3 confuses the signs inside the factors with the signs of the solutions.

The quadratic formula gives x=(2±√(4+4))/2=(2±2√2)/2=1±√2. The option 2±2√2 comes from forgetting to divide by 2 at the end. Carry out the simplification √8=2√2 and the cancelling carefully. Substituting x=1+√2 gives (1+√2)²−2(1+√2)−1=0, which confirms it.

The cross method gives (2x−1)(x+2)=0. From 2x−1=0 we get x=1/2, and from x+2=0 we get x=−2. Note that the factor (x+2) yields the solution −2, with the opposite sign. The option x=−1/2, 2 reverses both signs.

The condition for a repeated root is D=0. From D=16−4k=0 we get k=4. Then x²+4x+4=(x+2)²=0 with the repeated root x=−2. The option k=16 comes from writing D=16−k and forgetting the factor 4.

The condition is D>0. From D=9−4k>0 we get 4k<9, that is k<9/4. Equality (k=9/4) gives a repeated root, so it is excluded. The option k>−9/4 appears if the sign is mistaken and D is written as 9+4k.

For x²−4x+5=0 the discriminant is D=16−20=−4<0, so there is no real root and there are no common points. On the graph the vertex (2, 1) lies above the x-axis and the parabola opens upward, so the whole curve stays clear of the axis. A parabola and the x-axis can never have 3 points in common.

Set y=0 and solve x²+2x−3=0. From (x+3)(x−1)=0 we get x=−3, 1. The two numbers with product −3 and sum +2 are +3 and −1, and the factor (x+3) yields the solution −3. The option x=3, −1 reverses the signs.

The condition for tangency is D=0. From D/4=k²−(k+2)=k²−k−2=(k−2)(k+1)=0 we get k=2, −1. When k=2 we get x²−4x+4=(x−2)², and when k=−1 we get x²+2x+1=(x+1)², so both really do touch the axis.

Eliminating y gives x²+1=2x+4, which rearranges to x²−2x−3=0. From (x−3)(x+1)=0 we get x=−1, 3. The discriminant D=4+12=16>0 confirms that there really are 2 common points. Watch the signs when moving terms across.

The condition is D≧0. From D/4=(a+1)²−4≧0 we get (a+1)²≧4, so a+1≦−2 or a+1≧2, hence a≦−3 or a≧1. The option −3≦a≦1 takes the inside by mistaking the direction of the inequality. A repeated root is also a real root, so equality is included.

The roots of x²−5x+6=0 are x=2, 3. An upward parabola lies below the x-axis between the two roots, so the answer is 2<x<3. Taking the outside confuses this with the >0 case. Substituting x=2.5 gives 6.25−12.5+6=−0.25<0, which holds and confirms that the inside is right.

The roots of x²−x−6=(x+2)(x−3)=0 are x=−2, 3. For >0 the graph is above the x-axis, that is on the outside, so the answer is x<−2 or x>3. The interval −2<x<3 is the inside, the solution of <0. Note too that the factor (x+2) yields the root −2, so watch the signs.

The roots of x²−4=(x+2)(x−2)=0 are x=±2. For ≦0 we take the inside, and since equality is allowed the endpoints are included: −2≦x≦2. The option −2<x<2 drops the equality. At x=±2 the left side is 0, which satisfies ≦0, so the endpoints belong to the solution.

The cross method gives 2x²−5x+2=(2x−1)(x−2)=0 with roots x=1/2, 2. For <0, with no equality, we take the inside: 1/2<x<2. Substituting x=1 gives 2−5+2=−1<0, which holds and confirms the inside.

Here x²−6x+9=(x−3)², so the graph touches the x-axis at x=3. The expression (x−3)² is 0 only at x=3 and positive everywhere else, so the solution is every real number except 3. Saying "all real numbers" would wrongly include x=3, where the left side is 0.

Here D=4−12=−8<0, so the upward parabola has no point in common with the x-axis and always stays above it. The solution is therefore all real numbers. Completing the square as y=(x+1)²+2≧2>0 confirms this. Reading D<0 and mechanically answering "no solution" confuses this with a <0 inequality.

Multiply both sides by −1 and reverse the inequality sign to get x²−4x+3≦0. The roots of (x−1)(x−3)=0 are x=1, 3, and for ≦0 we take the inside, so 1≦x≦3. Forgetting to reverse the sign after multiplying by −1 leads to answering with the outside. At x=2 we get −4+8−3=1≧0, which also shows the inside is right.

Solving x²−2x−2=0 with the quadratic formula gives x=(2±√12)/2=1±√3. For <0 we take the inside, so 1−√3<x<1+√3. The procedure is the same even when the endpoints are irrational; the key is to carry out the simplification √12=2√3 and the cancelling correctly.

The condition for an upward parabola to stay above the x-axis everywhere is D<0. From D/4=k²−(k+6)=k²−k−6=(k−3)(k+2)<0 we get −2<k<3. Taking the outside would mean D>0, that is crossing the axis at 2 points, the opposite situation. At k=3 the curve touches the axis and equals 0 at x=−3, breaking the strict inequality, so the endpoints are excluded.

The first gives (x−5)(x+1)≦0, that is −1≦x≦5. The second gives x(x−2)>0, that is x<0 or x>2. The common part is −1≦x<0 together with 2<x≦5, so the integers are −1, 3, 4 and 5, four in all. The values 0 and 2 fail the second condition, since the left side is 0 there and not >0, so they are not counted.