Question

Y=x^2+px+g M(2;1) N(1;-2) найти p g

Answer 1

• т.к.M(2;1)∈графику ф-и,имеем 2²+2p+g=1
• N(1;-2)∈ГРАФИКУ,то  1+p+g= - 2
• 2p+g= - 3 и р+g= -3 это система,вычитая из ур-ия1  уравнение 2,получим p=0,тогда g= - 3
• y=x² - 3

Answer 2

To find the values of p and g, we can use the coordinates of points M(2, 1) and N(1, -2) on the quadratic equation y = x^2 + px + g. Since these points lie on the curve, we can substitute their coordinates into the equation and form a system of equations to solve for p and g.

Step 1: Substitute the coordinates of point M(2, 1) into the equation: 1 = (2)^2 + 2p + g 1 = 4 + 2p + g

Step 2: Substitute the coordinates of point N(1, -2) into the equation: -2 = (1)^2 + p + g -2 = 1 + p + g

Now we have a system of two equations with two variables (p and g):

1 = 4 + 2p + g ...(Equation 1) -2 = 1 + p + g ...(Equation 2)

To solve this system, we can use the method of substitution or elimination. Let's use the elimination method:

Step 3: Subtract Equation 2 from Equation 1 to eliminate g:

(1) - (-2) = (4 + 2p + g) - (1 + p + g) 3 = 3p + 3

Step 4: Now, isolate p by dividing both sides by 3:

3p = 3 p = 3/3 p = 1

Step 5: Now that we have the value of p, we can find g by substituting it back into either Equation 1 or Equation 2. Let's use Equation 1:

1 = 4 + 2(1) + g 1 = 4 + 2 + g 1 = 6 + g

Step 6: Finally, isolate g by subtracting 6 from both sides:

g = 1 - 6 g = -5

So, the values of p and g are: p = 1 g = -5