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?
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I believe you are missing some information. If the axis of symmetry is $x=2$ that would mean that $h=2$. Then $(3,7)$ gives us that $7=1+k\implies k=6$ but from $(4,11)$ we get that $11=4+k\implies k=7$. This is impossible. – John Douma Nov 13 '16 at 05:57
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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.
Muhammad Wasi
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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
$$ h = \frac{3}{2} , ~k= \frac{19}{4} $$
$$ \therefore y=(x-\frac{3}{2})^2 + \frac{19}{4} $$
bigfocalchord
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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 $(0,11)$.
John Douma
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