What is the range of the function f(x) = 3x2 + 6x – 8?

Mathematics

1 answer:

Fudgin [204]3 years ago

3 0

Answer:

Range → {y| y ≥ -11}

Step-by-step explanation:

Range of a function is the set of of y-values.

Given function is,

f(x) = 2x² + 6x - 8

By converting this equation into vertex form,

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

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

= $3[(x+1)^2-\frac{11}{3}]$

= $3(x+1)^2-11$

Vertex of the parabola → (-1, -11)

Therefore, range of the function will be → y ≥ -11

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Sindrei [870]

Answer:

Option A is your answer

3 0

3 years ago

Given: △ACM, m∠C=90°, CP ⊥ AM

larisa86 [58]

Answer: The answer is $3\dfrac{4}{7}.$

Step-by-step explanation: Given in the question that ΔAM is a right-angled triangle, where ∠C = 90°, CP ⊥ AM, AC : CM = 3 : 4 and MP - AP = 1. We are to find AM.

Let, AC = 3x and CM = 4x.

In the right-angled triangle ACM, we have

$AM^2=AC^2+CM^2=(3x)^2+(4x)^2=9x^2+16x^2=25x^2\\\\\Rightarrow AM=5x.$

Now,

$AM=AP+PM=AP+(AP+1)\\\\\Rightarrow 2AP=AM-1\\\\\Rightarrow 2AP=5x-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(A)$

Now, since CP ⊥ AM, so ΔACP and ΔMCP are both right-angled triangles.

So,

$CP^2=AC^2-AP^2=CM^2-MP^2\\\\\Rightarrow (3x)^2-AP^2=(4x)^2-(AP+1)^2\\\\\Rightarrow 9x^2-AP^2=16x^2-AP^2-2AP-1\\\\\Rightarrow 2AP=7x^2-1.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(B)$