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Tanzania [10]
4 years ago
5

Given a quadratic function, f(x) = ax2 + bx + c has a positive leading coefficient and the vertex that has a positive y-coordina

te. Determine the number of real zeros of the function
A. 2 real zeros

B. 1 real zero

C. 1 real zero and 1 imaginary zero

D. 2 imaginary zeros
Mathematics
2 answers:
marysya [2.9K]4 years ago
8 0
The correct answer is A. 2 real zeros.
Komok [63]4 years ago
6 0
As it has a positive leading coefficient the parabola will open upwards.

Also, as the y coordinate of the vertex is positive then the graph will not intersect the x axis at any point(s) ,  so there  can be no real zeroes.

D is the correct choice
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5 Victor knows that 2 + 7 = 7+2 and 2•7= 7•2
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The graph of f(x) = x2 is translated to form g(x) = (x – 5)2 + 1. On a coordinate plane, a parabola, labeled f of x, opens up. I
Firdavs [7]

The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).

<h3>How to analyze quadratic equations</h3>

In this question we have a graph of a <em>quadratic</em> equation translated to another place of a <em>Cartesian</em> plane, whose form coincides with the <em>vertex</em> form of the equation of the parabola, whose form is:

g(x) = C · (x - h)² - k     (1)

Where:

  • (h, k) - Vertex coordinates
  • C - Vertex constant

By direct comparison we notice that (h, k) = (5, 1) and C = 1. Now we proceed to check if the points (x, y) = (2, 10) and (x, y) = (8, 10) belong to the parabola.

x = 2

g(2) = (2 - 5)² + 1

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g(8) = (8 - 5)² + 1

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The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).

To learn more on parabolae: brainly.com/question/21685473

#SPJ1

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