We can find the vertex coordinates for this quadratic equation using the next formulas:
[tex]x_v=-\frac{b}{2a},y_v=c-\frac{b^2}{4a}[/tex]
Where the values for a, b, c are obtained by looking at the values of these coefficients in the general quadratic expression:
[tex]ax^2+bx+c[/tex]
Then, we have that:
a = 1, b = -3, c = -10.
Then, to find the x-coordinate of the quadratic equation, we have:
[tex]x_v=-\frac{-3}{2\cdot1}=\frac{3}{2}\Rightarrow x_v=\frac{3}{2}=1.5[/tex]
To find the y-coordinate, we have:
[tex]y_v=-10-\frac{(-3)^2}{4\cdot1}=-10-\frac{9}{4}=\frac{4\cdot(-10)-9}{4}=\frac{-40-9}{4}=-\frac{49}{4}=-\frac{48}{4}-\frac{1}{4}[/tex]
Then:
[tex]y_v=-\frac{48}{4}-\frac{1}{4}=-12-\frac{1}{4}=-12\frac{1}{4}=-12.25[/tex]
Therefore, the value for the y-coordinate is -12.25 (in decimal form).
In summary, we have that the vertex coordinates for the quadratic equation x²-3x-10 are (1.5, -12.25).
