Standard equations and simple properties of a Parabola

For a parabola in vertex form \(y=(x-h)^2+k\), the axis of symmetry is \(x=h\). Here, \(y=(x-(-2))^2-7\), so \(h=-2\), giving the axis \(x=-2\). The value \(y=-7\) is the vertex's y-coordinate, not the axis of symmetry. Exam tip: reverse the sign of the constant appearing with \(x\) inside the square to identify \(h\).

In the vertex form \(y=a(x-h)^2+k\), the axis of symmetry is \(x=h\). Here, \(x+3=x-(-3)\), so \(h=-3\) and the axis is \(x=-3\). The value \(y=7\) is the y-coordinate of the vertex, not the axis of symmetry. Exam tip: rewrite the squared term as \((x-h)^2\) and identify \(x=h\) directly.

Every point on the \(y\)-axis has \(x=0\). Thus, \(y=(0+1)^2-9=1-9=-8\), so the point of intersection is \((0,-8)\). Option B is the vertex of the parabola, not its \(y\)-intercept. Exam tip: To find a graph’s \(y\)-intercept, substitute \(x=0\).

For the quadratic, \(a=2\) and \(b=-12\). The x-coordinate of the vertex is \(x=-\frac{b}{2a}=\frac{12}{4}=3\). Substituting \(x=3\) gives \(y=2(3)^2-12(3)+13=-5\), so the vertex is \((3,-5)\). Option B has the wrong sign for the x-coordinate, while option D has the wrong sign for the y-coordinate. Exam tip: for \(y=ax^2+bx+c\), first find the vertex’s x-coordinate using \(-\frac{b}{2a}\).

Since \((x-3)^2\ge 0\), we have \(y=(x-3)^2+5\ge 5\). The \(x\)-axis is defined by \(y=0\), so the parabola never reaches or intersects it; hence, there are 0 intersections. Its vertex is \((3,5)\), which lies above the \(x\)-axis. Exam tip: To find intersections with the \(x\)-axis, set \(y=0\) and count the real solutions.