Finding the y-intercept

Question

The graph of the quadratic function y=(x-h)²+k passes through (3,7) and (4,11). The axis of symmetry is the graph is x=2. How can I find the y -intercept of the graph？ I know that it is the value of k (the minimum value of the function) but how could I find it? Can anyone please teach me?

Answer 1

Consider

$$ y=(x-h)^2+k $$

$$ \text{Passes through}~~ (3,7)~~ \text{and}~~ (4,11) $$

$$ \Leftrightarrow 7=(3-h)^2+k ~~~\text{and} ~~~ 11=(4-h)^2+k $$

Solving these 2 simultaneously

$$ h = \frac{3}{2} , ~k= \frac{19}{4} $$

$$ \therefore y=(x-\frac{3}{2})^2 + \frac{19}{4} $$

Answer 2

Y-intercept is the value of y at which graph touches y-axis(x=0). In the given case, y-intercept can be k only if h=0. You can compute it by puttting x=0 in the equation.

Answer 3

A quadratic equation describes a parabola. If the axis of symmetry is at $x=2$ then the $y$ value corresponding to $x=0$ is equal to the $y$ value at $x=4$. Therefore, the $y$ intercept is at $(0,11)$.