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vredina [299]
1 year ago
9

PLEASEEEEEE HELPPPPPPPPP!!!!!!!!!!!!!!!!! ON ASSESSMENTTTT PRACTICEEEE ITS VERY URGENT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Find the co

ordinate of the vertex for each of the following quadratics given below.

Mathematics
1 answer:
adelina 88 [10]1 year ago
8 0

In order to determine the vertex of the given functions, consider that the general form of a quadrati function is:

y = ax² + bx + c

The value of x for the vertex is given by:

x = -b/2a

The value for y, based on the previous values of x, is the y value of the vertex. Use the previous expression to find the vertices:

1. y = x² + 8x - 6

x = -8/2(1) = -4

y(-4) = (-4)² + 8(-4) - 6 = 16 - 32 - 6 = -22

Hence, the vertex is (-4,-22)

2. y = -4x² - 24x - 5

x = -(-24)/2(-4) = -6

y(-6) = -4(-6)² - 24(-6) - 6 = -144 + 144 - 6 = -6

Hence, the vertex is (-6,-6)

3. y = 2x² - 3x + 7

x = -(-3)/2(2) = 3/4 = 0.75

y(3/4) = 2(3/4)² - 3(3/4) + 7 = 18/16 - 9/4 + 7 =5.875

Hence, the vertex is (0.75 , 5.875)

4. y = -x² + 5x - 6

x = -5/2(-1) = 5/2 = 2.5

y(5/2) = -(5/2)² + 5(5/2) - 6 = 0.25

Hence, the vertex is (2.5 , 0.25)

5. y = 1/2 x² + 6x - 5

x = - 6/(2(1/2)) = -6

y(-6) = 1/2 (-6)² + 6(-6) - 5 = -23

Hence, the vertex is (-6 , -23)

6. y = 4x² + 7

x = -0/2(4) = 0

y(0) = 4(0) + 7 = 7

Hence, the vertex is (0 , 7)

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