A quadratic function is of the form f(x)=ax^{2}+bx+c and when we

yapafw

yapafw

Answered question

2021-12-03

A quadratic function is of the form f(x)=ax2+bx+c and when we set f(x), or y, equal to zero we call it
a quadratic equation ax2+bx+c=0. When we set y equal to zero and solve for x we are finding the x-
intercepts. Here are three quadratic equations:
y=x22x1
Using your graphing calculator find the x-intercepts, y-intercepts, and the maximum or minimum values of the quadratic equations.
Confirm your answers using the Quadratic Formula program.
Last, since your calculator only gives real number approximate answers for the zeroes solve these by hand to provide us with exact answers (might be complex numbers).

Answer & Explanation

Pulad1971

Pulad1971

Beginner2021-12-04Added 22 answers

The given quadratic equation is:
y=x22x1
First finding the intercepts of the equation.
For the x-intercept, put y=0, we have:
x22x1=0
Using quadratic formula for the roots of the equation.
x=2±(2)24×1×12
x=2±82
x=2(1±2)2
x=1±2
x=1+2
x=1+1.41
x=2.41
Also,
x=12
x=11.41
x=0.41
Thus, x-intercepts are 2.41 and -0.41.
Now, for y-intercepts, put x=0, we get:
y=(0)22(0)1
y=1
Thus,y-intercept is -1.
Now, using the formula for finding the maximum or minimum value of the given quadratic equation.
The formula is:
b2a
In the given quadratic equation, we have:
a=1,b=2,and c=1.
Thus,
x=(2)2(1)
x=22
x=1
Since ais positive ie., 1
Therefore, the quadratic equation has minimum value and it will be:
f(b2a)f(1)
f(1)=(1)22(1)1
f(1)=121
f(1)=2
Hence, -2 is the minimum value of the given quadratic equation at (1,-2)
Thus, the graph of the function with labelled values of intercepts and point of minimum value will be.

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