Simplify: 64r^8 8r^2 8r^4 32r^2 32r^4
1 answer:
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Answer:
The solutions are x = -9 , x = -5
Step-by-step explanation:
* Lets find the vertex of the parabola
- In the quadratic equation y = ax² + bx + c, the vertex of the parabola
is (h , k), where h = -b/2a and k = f(h)
∵ The equation is y = x² + 14x + 45
∴ a = 1 , b = 14 , c = 45
∵ h = -b/2a
∴ h = -14/2(1) = -14/2 = -7
∴ The x-coordinate of the vertex of the parabola is -7
- Lets find k
∵ k = f(h)
∵ h = -7
- Substitute x by -7 in the equation
∴ k = (-7)² + 14(-7) + 45 = 49 - 98 + 45 = -4
∴ The y-coordinate of the vertex point is -4
∴ The vertex of the parabola is (-7 , -4)
- Plot the point on the graph and then find two points before it and
another two points after it
- Let x = -9 , -8 and -6 , -5
∵ x = -9
∴ y = (-9)² + 14(-9) + 45 = 81 - 126 + 45 = 0
- Plot the point (-9 , 0)
∵ x = -8
∴ y = (-8)² + 14(-8) + 45 = 64 - 112 + 45 = -3
- Plot the point (-8 , -3)
∵ x = -6
∴ y = (-6)² + 14(-6) + 45 = 36 - 84 + 45 = -3
- Plot the point (-6 , -3)
∵ x = -5
∴ y = (-5)² + 14(-5) + 45 = 25 - 70 + 45 = 0
- Plot the point (-5 , 0)
* To solve the equation x² + 14x + 45 = 0 means find the value of
x when y = 0
- The solution of the equation x² + 14x + 45 = 0 are the x-coordinates
of the intersection points of the parabola with the x-axis
∵ The parabola intersects the x-axis at points (-9 , 0) and (-5 , 0)
∴ The solutions of the equation are x = -9 and x = -5
* The solutions are x = -9 , x = -5
