What is an equation of the axis of symmetry for the parabola whose equation is y=2x^2-3x+4

Mathematics

1 answer:

krek1111 [17]3 years ago

4 0

Answer:

x = 3/4 or x = 0.75

Step-by-step explanation:

y=2x²-3x+4

y'=4x-3 = 0

x = 3/4 .... answer

x = - b/2a ...y=ax²+bx+c

x = - (-3)/(2*2) = 3/4

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Kitty [74]

Answer: (-2, 5) and (2, -3)

Step-by-step explanation:

Graph the line y = -2x + 1 (which is in y = mx + b format) by plotting the y-intercept (b = 1) on the y-axis and then using the slope (m = -2) to plot the second point by going down 2 and right 1 unit from the first point:

y - intercept = (0, 1) 2nd point = ( -1, 1).

Graph the parabola y = x² - 2x - 3 by first plotting the vertex and then plotting the y-intercept (or some other point):

$y = x^2-2x-3\quad \rightarrow \quad a=1,\ b=-2,\ c=-3\\\\\text{axis of symmetry:}\ x = \dfrac{-b}{2a}\ \longrightarrow \ x=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\\text{y-value of vertex:}\ f(1) = (1)^2-2(1)-3\quad \longrightarrow \quad y = 1 - 2 - 3=-4\\\\\text{y-intercept:}\ f(0)= (0)^2-2(0)-3\ \longrightarrow \ y=0 - 0 - 3 = -3 \\$