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liq [111]
2 years ago
10

Help with Another Pre Calculus Question

Mathematics
1 answer:
swat322 years ago
8 0

The last  equation h(x) = 2x+4/x+1.

<h3>What is a graph?</h3>

The pictorial representation of quantities and how the increase or decrease in one affects the other.

Analysis:

The graph cuts the x-axis and y-axis at (-2, 0) and(0, -4) respectively.

gradient of the curve = y2 - y1/x2 - x1 = -4-0/0--2 = -4/2 = -2.

The equation with a gradient equal to -2 is, h(x) = 2x+4/x+1

differentiating the equation by quotient rule,

u = 2x+4, du/dx = 2, v = x+1 dv/dx = 1

dy/dx = \frac{(x+1)(2) - (2x+4)(1)}{(x+1)^{2} } = \frac{-2}{(x+1)^{2} }

When x = -2, dy/dx = -2

In conclusion, 2x+4/x+1 is the equation for the curve above.

Learn more about parabolic curves: brainly.com/question/12896871

#SPJ1

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The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

Step-by-step explanation:

we know that

The equation of a vertical parabola in vertex form is equal to

f(x)=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex of the parabola

If a > 0 the parabola open upward and the vertex is a minimum

If a < 0 the parabola open downward and the vertex is a maximum

In this problem

The coefficient a must be positive, because we need to find a minimum

therefore

Check the option C and the option D

Option C

we have

f(x)=\frac{1}{2}x^{2}-4x+5

Convert to vertex form

f(x)-5=\frac{1}{2}x^{2}-4x

Factor the leading coefficient

f(x)-5=\frac{1}{2}(x^{2}-8x)

f(x)-5+8=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x^{2}-8x+16)

f(x)+3=\frac{1}{2}(x-4)^{2}

f(x)=\frac{1}{2}(x-4)^{2}-3

The vertex is the point (4,-3) ( is a minimum)

therefore

The graph of the function f(x)=\frac{1}{2}x^{2}-4x+5 has a minimum located at (4,-3)

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