Question

The y-coordinate of the vertex of the parabolic graph of
f(x) = ax2 + bx + c
is

Answer 1

hello,

y-coordinate is found by computing f(0)=c

so this is the point (0,c)

the vertex is found for x = -b/(2a)

$f(\dfrac{-b}{2a})=\dfrac{ab^2}{4a^2}-\dfrac{b^2}{2a}+c\\\\=\dfrac{b^2-2b^2+4ac}{4a}\\\\=-\dfrac{b^2-4ac}{4a}$

So this is the point ( -2(2a), -(b^2-4ac)/(4a) )

thanks

Answer 2

The vertex of the parabolic graph is, $\rm\left ( -4a, \dfrac{-b^2+4ac}{4a} \right )$

Given that,

$\rm f(x) = ax^{2} +bx+c$

We have to determine,

The y-coordinate of the vertex of the parabolic graph?

According to the equation,

The vertex of a parabola is the point at which the parabola passes through its axis of symmetry.

In case the coefficient of the x² term is positive, and then the vertex will be located at the lowest point on the graph, the point at the base of the “U”-shape.

On the contrary, if the coefficient of the x² term is negative, the vertex will be located at the highest point on the graph, at the top of the “U”-shape.

Therefore,

The vertex of the parabola of the given quadratic equation,

$\rm f(x) = ax^{2} +bx+c$

The co-ordinate is founded when x = 0,

$\rm f(x) = ax^{2} +bx+c\\\\\rm f(0) = a(0)^{2} +b(0)+c\\\\\rm f(0) = c$

Then, The vertex is found at point (0, c) at x = -b\2a,

$\rm f \left (\dfrac{-b}{2a }\right) = a\left (\dfrac{-b}{2a }\right) ^2 + b \left (\dfrac{-b}{2a }\right) +c\\\\f \left (\dfrac{-b}{2a }\right) = a \left (\dfrac{b^2}{4a^2 }\right)- \left (\dfrac{b^2}{2a }\right) +c\\\\f \left (\dfrac{-b}{2a }\right) = \left (\dfrac{b^2}{4a }\right)- \left (\dfrac{b^2}{2a }\right) +c\\\\f \left (\dfrac{-b}{2a }\right) = \dfrac{b^2-2b^2+4ac}{4a}\\\\f \left (\dfrac{-b}{2a }\right) = \dfrac{-b^2+4ac}{4a}$

Hence, The vertex of the parabolic graph is, $\rm\left ( -4a, \dfrac{-b^2+4ac}{4a} \right )$.

For more details refer to the link given below.

https://brainly.com/question/1254213