What is the vertex f(x)= 5x^2+20x-16

1 answer:

3 0

F(x) = 5x² + 20x - 16

y = 5(x² + 4x) - 16
y + 16 = 5(x² + 4x + __)

perfect square:
x² ⇒ x * x
4x ⇒ 2*2x
2² = 4

y + 16 + 4 = 5(x² + 4x + 4)
y + 20 = 5(x+2)²

(x+2)² ⇒ (x+2)(x+2) = x(x+2) + 2(x+2) = x² + 2x + 2x + 4 = x² + 4x + 4

y = 5(x+2)² - 20

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50 points! What is the value of x? Enter your answer in the box. x =

frosja888 [35]

Answer:

x = 3

Hope this helps!

Step-by-step explanation:

ΔABC is a 45°, 45°, 90° triangle. The longest side ( 6 $\sqrt{2}$ ) can be $x\sqrt{2}$ while AB and CB are x. Because 6 $\sqrt{2}$ is equal to $x\sqrt{2}$ , you can see that x is equal to 6. Both AB and CB are equal to 6. ΔBDC ( CDB ) is a right triangle ( 30°, 60°, 90° ) so, BD = x, CB = 2x, and CD = $x\sqrt{3}$ . CB is equal to 6 so 2x = 6, x = 3.