Who has a question of a primary function and a quadratic function with ten answers! Ten to it

1.The parabola y=ax2+bx+c passes through the point a(-1,2)b(2,-1) and intersects the y-axis at the point m

Question: Find the answer to the coordinates of the point on the parabola y=ax2-bx+c-1 where the abscissa is equal to the ordinate. Because the parabola y=ax2-bx+c is symmetrical with y=ax2+bx+c with respect to the y-axis.

Crossing point A (-1,2).

b(2,-1), so point(1,2).

2,-1) is on the parabola y=ax2-bx+c, and the parabola y=ax2-bx+c-1 is the image of the parabola y=ax2-bx+c that is translated downwards by one unit.

So the points (1,1) and points (-2,-2) are the points 2If the line y=x-1 intersects the parabola y=x2+5x+a2, then their common point must be in what quadrant?

Answer; Third Quadrant.

The straight line y=x-1 intersects the parabola y=x2+5x+a2, so x-1=x2+5x+a2

i.e. x 2 + 4 x + a 2 + 1 = 0

It can be seen according to the equation.

If the equation has roots.

The root must be less than 0

and y=x-1

So y is also less than 0

So the intersection is in the third quadrant.

The straight line y=(-4 3)x+4 intersects the x-axis at the point a, and the y-axis intersects at the point c, and the image of the known quadratic function passes through the points a, c, and b(-1,0).

Question: There are two moving points d and e starting from point o at the same time, where point d moves along the line oac according to the route of o-a-c at a speed of 3 2 units of length per second, and point e moves along the line oca according to the route of o-c-a at a speed of 4 UNIT LENGTHS PER LINGYE SEARCH SECOND, WHEN D AND E ARE SENSITIVE, THEY BOTH STOP MOVING, LET D AND E HAVE THE SAME RULER DURATION FROM POINT O WHEN T SECONDS, THE AREA OF ODE IS S.

1.Is there a de oc in the process of movement of points d and e? If it exists, the value of t at this time is requested; If not, please state the reason;

2.Request the functional relation of s about t, and write the range of values of the independent variable t;

3.Let s1 be the maximum value of the function s in 2, then s1=?

1, =a 4(a 2)=a 4a 8=(a 2) 4>0, so there are two intersections with the x-axis;

2. The distance between these two intersections =|x1－x2|=(x1 x2) 4x1x2=( a) 4(a 2)=(a 2) 4, then the minimum distance between these two intersections is 2, where a=2.

The parabola passes through the three points a(-1,0)b(3,0)c(0,3), and the parabola is set to y=ax 2+bx+c

Substitute the coordinates of a, b, and c into it.

c=3a-b+c=0

9a+3b+c=0

Get a=-1, b=2, c=3

1) The parabolic analytic formula is y=-x 2+2x+3(2) when x=m, y=-m 2+2m+3ob=oc=3, and obc is an isosceles triangle.

dm=db=3-m

mn=-m 2+2m+3-(3-m)=-m 2+3m(3) Area of BNC = mn*od 2+mn*db 2=mn*ob 2=(-m 2+3m)*3 2

(3 2)*(m-3 2) 2+(3 2) 3When m=3 2, the area of BNC is the largest.

1) Substituting the points a(-1,0), b(3,0), and c(0,3) into y=ax 2+bx+c obtains: a=-1, b=2, c=3, so the analytic formula of the parabola is: y=-x 2+2x+3;

Solution: Let f(x)=ax 2+bx+c, (a is not equal to 0).

Since f(-3 2 +x) = f(-3 2 -x), then, f(-3 2 +(x+3 2))) = f(-3 2 -(x+3 2)).

That is, f(x)=f(-x+3), substituting the function gives ax 2+bx+c=a(-x+3) 2+b(-x+3)+c

(2b+6a)x-(9a+3b)=0, that is, 2b+6a=0 and 9a+3b=0, b=-3a

ax 2+bx+c=0,x1+x2=-b a=3,x1-x2=7,x1*x2=c a, x1=5,x2=-2,x1*x2=-10, the equation can be converted to a(x-5)(x+2)=0, that is, ax 2-3ax-10a=0, if there are no other conditions, a can take any real value that is not 0, so there are many correct answers, of course, f(x)=-4x 2+12x+40 is also one of them, when a=- 4 o'clock results. Therefore, f(x) = ax 2-3ax-10a(a is not equal to 0), here is just the solution method, to do the function problem to use both the displayed condition and the implied condition.

Answer: y=3x 2-mx+3 and the intersection of y-axis b(0,3);

When intersecting with the x-axis, the quadratic equation should be solved: 3x 2-mx+3=0, and the problem is set to intersect with the positive semi-axis of the x-axis, indicating that the real root of the equation exists, the discriminant formula > 0, the sum of the two roots is m 3, and the product of the two roots is 1, indicating that both roots are positive, m>0, and when the discriminant formula m 2-36=0, m=6, x=1, so a(1,0);

In the discriminant formula m 2-36>0, i.e., m>6, there are two solutions, that is, there are two intersections between the function image and the x-axis positive semi-axis:

a((m- (m2-36)) 6,0), or a((m+ (m2-36)) 6,0) (both with parameter m).

For such questions, consider using the special value method, such as finding the answer when y=0 or x=0.

y=3x^2-mx+3

When x=0, y= 3=》y-axis intersects at point b, b coordinates are (0,3), when y=0, 3x 2-mx+3=0=" x = (m+ (m2-36)) 6 or x = (m- (m2-36)) 6

From the root number, we get m>=6 or m<=-6

And because the function intersects with the x positive semi-axis at point a.

When m<=-6, it is not in line with the topic (discarded).

The coordinates of point A are ((m+ (m2-36)) 6,0).or ((m- (m2-36)) 6,0) m belongs.

y=3x^2-mx+3

x=0y= 3

The y-axis intersects at point b, => b=(0,3).

y=03x^2-mx+3=0

x = (m+ m 2-36) 6 or (m- m 2-36) 6x positive semi-axis intersection at point a, => a= ((m+ m 2-36) 6,0) or ((m- m 2-36) 6,0).

(1) Substituting x=1, y=2 into the equation y=ax squared + bx obtains: a+b=2;①

Substituting x=2, y=2+6=8 into the equation y=ax squared + bx yields: 4a+2b=8, i.e., 2a+b=4;②

Get: a=2, substitute for b=0

So, the functional relationship between y and x is: y=2x 2;

2）w=33x-y-100=-2x^2+33x-100；

3) w=-2x 2+33x-100, a quadratic function with the opening downward, the axis of symmetry is x=33 4=, and the nearest positive integer to the axis of symmetry is x=8, so after eight months, the net return reaches the maximum;

The return on investment, that is, breakeven, is 0, that is, w==-2x 2+33x-100=0, collated: 2x 2-33x+100=0, multiplication of crosses: (2x-25) (x-4) = 0, x=4 or x=25 2, so, after four months, the investment can be recovered.

Note: The upstairs is wrong, and the title says that "the cumulative maintenance cost from the first month to the first month is y (ten thousand yuan)".

Note the word "cumulative", so when x = 2, y 2 6 8

I hope it can help you, if you don't understand, please hi me, I wish you progress in your studies!

Solution: (1) From the meaning of the question, a+b=2 and 4a+2b=6, a=b=1, so y=x +x

2) w=33x-y-100=-x +32x-100(3)w=-(x-16) +156, so when x=16, w=156 is the maximum.

From w>=0, the investment can be recovered after 4 months.

(1) 2=a+b, 6=4a+2b, a=1, b=1 y=x, squared+x

2) w = (33-1) x - 100-x squared.

3) According to the axis of symmetry formula -b 2a = 16, i.e. the net gain is the largest after 16 months!

0=32x-100-x squared, x is approximately equal to 4, so after 4 months, the investment can be recovered.