course mth 163
I chose Version #12 of the randomized problem set: It says Find the vertex and zeros of :y= .026t^2 + -1.2t + 41
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In order to find the vertex, we use the formula x= -b/(2a). We substitute the numbers in:
x = -(-1.2)/(2 * .026)
x = 1.2/(.052)
x = 23.08
Substitute this back in to obtain y coordinate:
y = .026* (23.08)^2 + (-1.2)(23.08) + 41
y = .026(532.6864) - 27.696 + 41
y = 13.8498464 + 13.304
y = 27.15
The VERTEX is: (23.08, 27.15)
To obtain the zeros for the formula we use:
x= [-b + (sqrt b^2 -4ac)]/2a & x= [-b - (sqrt b^2 -4ac)]/2a
Substitute in:
x= [-b + (sqrt b^2 -4ac)]/2a
x= [-(-1.2) + (sqrt (-1.8)^2 -4(.026)(41)]/2(.026)
x= [1.2 + (sqrt [(1.44) - 4.264]]/.052
x= [1.2 +(sqrt -2.824)]/.052
Upon seeing this you cannot get square root of a negative number so there are no zeros in this system. "
Very good.
The fact that the parabola opens upward and has a vertex above the x axis reinforces the fact that there are no zeros.