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Hint: You have a quadratic in the form \(\large y= ax^2 + bx + c\) The vertex is located *x-coordinate* at \(\large x = -\frac{b}{2a}\) Then simply plug that 'x' back into the original function to find the y-coordinate
Is it above the x axis?
Are you asking me or telling me? haha did you go through the process?
Asking. Duhhhh No....I kinda just graphed it
Lol well graphing works too! But yes, above the x-axis
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Thank youuuu! c;
Of course :)
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