$$ f(x)=x^{2}-10 x-56 $$ 1) What are the zeros of the function? Write the smaller $$ x $$ first, and the larger $$ x $$ second. smaller $$ x= $$ larger $$ x= $$ 2) What is the vertex of the parabola? 2

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1) To find the zeros of f(x) = x² - 10x - 56, we set the function equal to zero:

x² - 10x - 56 = 0

Using the quadratic formula, where a = 1, b = -10, and c = -56:

x = [ -b ± √(b² - 4ac) ] / (2a)
  x = [ 10 ± √(100 + 224) ] / 2
  x = [ 10 ± √324 ] / 2
  x = [ 10 ± 18 ] / 2

This gives us two solutions:

x = (10 - 18)/2 = -8/2 = -4  
  x = (10 + 18)/2 = 28/2 = 14

Smaller x = -4  
Larger x = 14

2) The vertex of a parabola given by f(x) = ax² + bx + c is found using:

x-coordinate: x = -b/(2a)
  For f(x) = x² - 10x - 56, a = 1 and b = -10, so:
  x = -(-10)/(2(1)) = 10/2 = 5

Now, substitute x = 5 back into f(x) to find the y-coordinate:

f(5) = 5² - 10(5) - 56  
    = 25 - 50 - 56  
    = -81

So the vertex is (5, -81).
1) The zeros of the function are -4 and 14. 2) The vertex of the parabola is at (5, -81).

\( f(x)=x^{2}-10 x-56 \) 1) What are the zeros of the function? Write the smaller \( x \) first, and the larger \( x \) second. smaller \( x= \) larger \( x= \) 2) What is the vertex of the parabola? 2

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