Quadratic relations

Question

Given y=-(x-5)^2-1 Find each of the following 

coordinates of vertex 

is the parabola maximum or min value 

the maximum or min value 

write the equation in standard form 

the y Intercept 

the x intercept 

 

Answer 1

 

y = (x - 5)² - 1

 

This equation is in vertex form ...  y = a(x - h)² + k ... the vertex is (h, k)

 

The vertex is (5, -1)

 

This parabola opens upward because "a" is positive ... a = 1 ....

Thus, the vertex is a minimum.

 

The minimum value is the y-coordinate of the vertex.

The minimum value is -1.

 

The standard form format of the equation is ... Ax + bx + C

 

    y = (x - 5)² - 1

    y = (x - 5)(x - 5) - 1

    y = x² - 10 x + 25 - 1

    y = x² - 10x + 24

 

The standard form of the equation is ... y = x² - 10x + 24

 

To find the y intercept, set x = 0 and solve for y

 

     y = x² - 10x + 24

     y = 0² -10(0) + 24

     y = 24

 

The y-intercept is (0, 24)

 

To find the x-intercept, set y = 0 and solve for x

 

      y = x² - 10x + 24

      0 = (x - 6)(x - 4)

      

      Use zero product property

 

      x - 6 = 0; x = 6

     

      x - 4 = 0; x = 4

 

Thus, the x-intercepts are (4, 0) and (6, 0)