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maths functions

Please Help Maths Functions class=

Sagot :

Answer:

The Co-ordinate of C is (4/3, -1/2)

Step-by-step explanation:

We have two equation one is of straight line equation which is:

                                              y=2x-3           (i)

Other equation is of quadratic function which is:

                                         y=-3x^2+5         (ii)

Put the value of y from equation (i) in equation (ii)

So, we have:

                                     2x-3=-3x^2+5

                                      3x^2+2x-8=0

By factorization:

                                    3x^2+6x-4x-8=0

                                   3x(x+2)-4x(x+2)=0

                                        (x+2)(3x-4)=0

              x+2=0                             ;                      3x-4=0

              x=-2                                ;                       x=4/3

Put first x=-2 in equation (i)

                                                y=2(-2)-3

                                                  y=-4-3

                                                    y=-7

Now Put x=4/3 in equation (i)

                                             y=2(4/3)-3

                                                 y=8/3-3

                                                   y=-1/2

So, we have two Order pair One is (-2 , -7) and Second one is (4/3 , -1/2)

Hence the Co-ordinate of C is:

                                               C=(4/3 , -1/2)

View image azamrasheed2310
semsee45

Answer:

Point C:  (3, 3)

Point D:  (3, -22)

Step-by-step explanation:

If the distance between points C and D is 25 units, the y-value of point D will be 25 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x-3[/tex]

[tex]\textsf{Equation 2}: \quad y-25=-3x^2+5[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x-3-25=-3x^2+5[/tex]

[tex]\implies 3x^2+2x-33=0[/tex]

[tex]\implies 3x^2-9x+11x-33=0[/tex]

[tex]\implies 3x(x-3)+11(x-3)=0[/tex]

[tex]\implies (x-3)(3x+11)=0[/tex]

[tex]\implies x=3, -\dfrac{11}{3}[/tex]

From inspection of the given graph, the x-value of points C and D is positive, therefore x = 3.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x-3 & =y\\2(3)-3 & =3\\ \implies & (3, 3)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -3x^2+5 & = y \\ -3(3)^2+5 & =-22\\ \implies & (3, -22)\end{aligned}[/tex]

Therefore, point C is (3, 3) and point D is (3, -22).

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