Answer:
y = (x -5)² + 3.
Step-by-step explanation:
Given : parabola with a vertex at (5,3).
To find : Which equation has a graph that is a parabola.
Solution : We have given vertex at (5,3).
Vertex form of parabola : y = (x -h)² + k .
Where, (h ,k ) vertex .
Plug h = 5 , k= 3 in vertex form of parabola.
Equation :y = (x -5)² + 3.
Therefore, y = (x -5)² + 3.
The nature of a graph, which has an even degree and a positive leading coefficient will be<u> up left, up right</u> position
<h3 /><h3>What is the nature of the graph of a quadratic equation?</h3>
The nature of the graphical representation of a quadratic equation with an even degree and a positive leading coefficient will give a parabola curve.
Given that we have a function f(x) = an even degree and a positive leading coefficient. i.e.
The domain of this function varies from -∞ < x < ∞ and the parabolic curve will be positioned on the upward left and upward right x-axis.
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The value of x in the given expression can be either -41 or 45. The correct option is B.
<h3>What is Absolute Value?</h3>
An absolute value of |x| {modulus of x} is the value of a real number x, the value we get is always a non-negative number, for example, |-5| will give 5, and also, |5| will give 5 as well.
Given the modulus function (1/5)|x - 4| - 3 = 6. Now, there will be two cases of this because the value of x can be either negative, x<0 or it can positive, x>0. Therefore, the given modulus function can be solved as shown below.
(1/5)(x-4) - 3 = 6
(x/5) - (4/5) - 3 = 6
(x - 4 - 15)/5 = 6
x - 4 -15 = 30
x = 30 + 15 + 4
x = 45
(1/5)(-x+4) - 3 = 6
(-x/5) + (4/5) - 3 = 6
(- x + 4 - 15)/5 = 6
- x + 4 -15 = 30
-x = 30 + 15 - 4
-x = 41
x = -41
Hence, the value of x in the given expression can be either -41 or 45.
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