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Brrunno [24]
3 years ago
8

Consider the quadratic function f(x) = x2 – 8x – 4. What is the value of the leading coefficient

Mathematics
2 answers:
jarptica [38.1K]3 years ago
7 0

Answer:

The answer is

the value of the leading coefficient is equal to 1

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form ax^{2} +bx+c=0 is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

f(x)=x^{2} -8x-4  

equate the function to zero

x^{2} -8x-4=0    

so

a=1\\b=-8\\c=-4

the value of the leading coefficient is equal to a

a=1


steposvetlana [31]3 years ago
5 0
Easey
f(x)=ax^2+bx+c
leading coefient is a
f(x)=1x^2-8x-4
leading coefient is 1
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3 years ago
Graph a parabola whose x-intercepts are at x=-3 and x=5 and whose minimum value is y=-4
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Answer:

(See explanation for further details)

Step-by-step explanation:

The standard equation of the parabola is:

y + 4 = C \cdot (x-k)^{2}

The formula is now expanded into a the form of a second-order polynomial:

y + 4 = C\cdot x^{2} -2\cdot C\cdot k \cdot x +C\cdot k^{2}

y = C\cdot x^{2} - (2\cdot C \cdot k) \cdot x + (C\cdot k^{2}-4)

The general equation of the second-order polynomial is:

x = \frac{2\cdot C \cdot k \pm \sqrt{4\cdot C^{2}\cdot k^{2}-4\cdot C\cdot (C\cdot k^{2}-4)}}{2\cdot C}

x = k \pm \frac{\sqrt{C^{2}\cdot k^{2}-C^{2}\cdot k^{2}+4\cdot C}}{C}

x = k \pm 2\cdot \frac{\sqrt{C}}{C}

x = k \pm \frac{2}{\sqrt{C}}

The equations to be solved are presented herein:

-3 = k -\frac{2}{\sqrt{C}}

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Now, the solution of the system is:

-3 +\frac{2}{\sqrt{C}} = 5 -\frac{2}{\sqrt{C}}

\frac{4}{\sqrt{C}} = 8

\sqrt{C} = \frac{1}{2}

C = \frac{1}{4}

k = 5 - \frac{2}{\sqrt{\frac{1}{4} }}

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The equation of the parabola is:

y = \frac{1}{4}\cdot (x-1)^{2} -4

Lastly, the graphic of the function is included as attachment.

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