hooglwired.blogg.se - Standard form quadratic

Now we see that the linear coefficient $b=-2ah$ must be positive (since $a$ is positive and $h$ is negative). How does this help with $a$, $b$, and $c$? Well, if we write $f(x)$ in the form $$f(x)=a(x-h)^2+k=ax^2-2ahx+(h^2+k),$$ we see that the two sets of coefficients are related by $b=-2ah$ and $c=h^2+k$. Next, by its location in the second quadrant we can also see that the coordinates of the vertex $(h,k)$ satisfy $h0$. Writing $f(x)=ax^2+bx+c$, we first see that since the given parabola is upward-facing, we must have $a>0$.

But as it turns out, we'll be able to reason whether each one of the coefficients is positive or negative. To ensure that $f(6)=-1$, we need to ensure that $a(6-4)^2-9=-1$, which forces $a=2$.ģ) Of course, without any scales on the axes, it will be impossible to figure out exact values for these coefficients.

Any such example takes the form $f(x)=a(x-4)^2-9$ for some constant $a$.

One example is $f(x)=(x+7)^2 = x^2+14x+49$. Once the quadratic is in standard form, the values of a a, b b, and c c can be found.

Transformational/graphical approach: The graph is a scaled reflection of the graph of $y=x^2$ over the $x$ axis, and so takes the form $y=ax^2$ for some negative value of $a$. Taking a single point on the graph, say $(5,-12.5)$, means we must have $-12.5=a(5^2)$, giving $a=-\frac$, $b=0$, and $c=0$.

We outline some of them here (which overlap heavily in places), applied to the top left graph, and then only give the final answers in the solution below: There are many possible approaches to solving each part of this problem, especially the first part. This task has students explore the relationship between the three parameters $a$, $b$, and $c$ in the equation $f(x)=ax^2+bx+c$ and the resulting graph.