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Algebra
Question

Again, we are solving for " $$+ C ^ { \prime \prime }$$ using the given values below, and replacing it into the new integrated function. $$\frac { d y } { d x } = 4 x + 2 , y ( 0 ) = 1$$

$$y = 2 x ^ { 2 } + 2 x + 3$$

$$y = 2 x ^ { 2 } + 2 x - 1$$

$$y = 2 x ^ { 2 } + 2 x + 1$$

$$y x ^ { 2 } + 2 x - 2$$

y=2x$$^{2}$$+2x+c
y=2x$$^{2}$$+2x+1