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1. The given function: f(x) = 2(x-4)²+7

a. Determine how it opens?
b. Find the Vertex?
c. Find the X-intercept?
d. Find the Y-Intercept?
e. Graph the function.

User Overgroove
by
7.7k points

1 Answer

4 votes

Answer:

a. Parabola opens upward

b. Vertex at (4, 7)

c. No x-intercept

d. y-intercept = 39 at point(0, 39)

e. Attached

Explanation:

The function is a quadratic function and the function graph represents a parabola

The equation of a parabola in vertex form is
y = a(x -h)² + k

The vertex is found at (h, k)

If a is positive, the parabola opens upward, if negative, the parabola opens downward.

Mapping the given function f(x) = y = 2(x - 4)² + 7 to the general equation we can see that
a = 2
h = 4
k = 7

a. Since a >0 the parabola opens downward

b. The vertex is at (h, k) => (4, 7)

c. The x-intercept can be found by setting y = 0 and solving for x
Setting y = 0 gives
0 = 2(x -4)² + 7

Switching sides,
2(x-4)² + 7 = 0

2(x-4)² = -7

(x-4)² = -7/2

(x - 4) =√(-7/2)

Since square roots of negative numbers are not real numbers, the parabola does not have an x-intercept

c. To find the y-intercept set x = 0 and solve for y
=> y = 2(0-4)² + 7

y = 2 (-4)² + 7

y = 2 x 16 + 7

y = 39

So y-intercept is at the point(0, 39)

e. Graph provided

1. The given function: f(x) = 2(x-4)²+7 a. Determine how it opens? b. Find the Vertex-example-1
User MrLeblond
by
8.2k points

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