Parabolas: Practice

Read the leading coefficient: here $a = -2$.

$a = -2 < 0 \;\Rightarrow\; \text{opens downward}$

The negative $a$ makes the parabola a downward hill. The $+3$ only lifts it, it does not change the direction.

Width is set by the size of the leading coefficient, $\lvert a \rvert$: the larger it is, the faster the arms climb and the narrower the parabola.

$\lvert 4 \rvert = 4 \text{ is the largest} \;\Rightarrow\; y = 4x^2 \text{ is narrowest}$

The others have $\lvert a \rvert = 1$, $\tfrac{1}{2}$, and $1$, all smaller than $4$.

In vertex form $y = a(x - h)^2 + k$, the vertex is $(h, k)$. Matching $y = (x - 4)^2 + 1$ gives $h = 4$ and $k = 1$.

$(h, k) = (4, 1)$

The number inside the parentheses is subtracted, so $x - 4$ means $h = +4$.

Set $x = 0$ to find where the graph crosses the y-axis. The $x^2$ and $x$ terms drop out, leaving the constant term $c = 9$.

$y = 2(0)^2 - 4(0) + 9 = 9$

So the y-intercept is $(0, 9)$.

The vertex is the turning point. For $y = x^2$ the smallest value of $y$ is $0$, reached when $x = 0$.

$x = 0 \;\Rightarrow\; y = 0^2 = 0$

So the vertex is $(0, 0)$, the origin, the lowest point of the upward bowl.

Setting $x = h$ makes the squared term $0$, so $y = k$, and that is the turning value.

$x = h \;\Rightarrow\; y = a(0)^2 + k = k$

So the vertex is $(h, k)$, read straight off vertex form.

The axis of symmetry is the vertical line $x = -\tfrac{b}{2a}$. For $y = x^2$ we have $a = 1$ and $b = 0$.

$x = -\frac{0}{2(1)} = 0$

So the axis is $x = 0$, the y-axis, which is the fold line where the left and right halves match.

When a parabola opens upward it is a valley, so the curve climbs on both sides of the turning point.

$a > 0 \;\Rightarrow\; \text{vertex is the lowest point}$

The vertex is the bottom of the bowl. For a downward parabola it would be the highest point instead.

Set $x = 0$. The squared and linear terms vanish, leaving the constant term $c = -2$.

$y = -(0)^2 + 5(0) - 2 = -2$

So the y-intercept is $(0, -2)$.

Compare with vertex form $y = a(x - h)^2 + k$: here $h = 1$ and $k = 3$.

$(h, k) = (1, 3)$

The $x - 1$ gives $h = 1$, and the $+3$ gives $k = 3$.

The point $(0, c)$ has $x = 0$, so it lies on the y-axis.

$x = 0 \;\Rightarrow\; y = c, \quad \text{the point } (0, c)$

A point where a graph crosses the y-axis is a y-intercept.

In the form $y = (x - 0)^2 - 9$ we have $h = 0$ and $k = -9$.

$(h, k) = (0, -9)$

There is no $x$ term, so the vertex stays on the y-axis, dropped $9$ units to $(0, -9)$.