Quadratic formula

Exercise

Given the quadratic polynomial in standard form \(4x^2+4x-7\)

(i) Find the roots of the polynomial using the quadratic formula.

(ii) Draw the x-intercepts and the axis of symmetry.

(iii) Find the y-value of the vertex by evaluating the polynomial.

(iv) Graph the parabola using three points.

Solution

(i) Quadratic formula

Step 1: Write coefficients

The coefficients are

\[a=4, \quad b=4,\quad c=-7\]

Step 2: Calculate discriminant

The discriminant is

\[\begin{align*} D&=b^2-4ac\\ &=(4)^2-4(4)(-7)\\ &= 16+112\\ &=128 \end{align*}\]

Step 3: Apply quadratic formula

Quadratic formula

\[\begin{gather*} -\frac{b}{2a} \pm \frac{\sqrt{D}}{2a}\\ = -\frac{(4)}{2(4)} \pm \frac{\sqrt{128}}{2(4)}\\ = - \frac{4}{8} \pm \frac{\sqrt{128}}{8}\\ =-\frac{1}{2} \pm \frac{\sqrt{64}\sqrt{2}}{8}\\ =-\frac{1}{2} \pm \frac{8\sqrt{2}}{8}\\ =-\frac{1}{2} \pm \sqrt{2} \end{gather*}\]

Roots \(x_1 =-\frac{1}{2}-\sqrt{2}, \quad x_2 =-\frac{1}{2}+\sqrt{2}.\)

(ii) x-intercepts and axis of symmetry

Step 1: x-intercepts

The x-intercepts are

\[(x_1,0),\quad (x_2,0)\]

which is

\[\Big(-\frac{1}{2}-\sqrt{2},0\Big),\quad \Big(-\frac{1}{2}+\sqrt{2},0\Big).\]

Step 2: Axis of symmetry

The axis of symmetry is the vertical line \(x=h\) where

\[\begin{align*} h &= -\frac{b}{2a}\\ &= -\frac{(4)}{2(4)}\\ &=-\frac{4}{8}\\ &=-\frac{1}{2}. \end{align*}\]

Axis of symmetry

\[x=-\frac{1}{2}.\]

(iii) y-value of vertex

Use \(4x^2+4x-7\) with \(x=-\frac{1}{2}\):

\[\begin{align*} k&=4\big(-\frac{1}{2}\big)^2+4\big(-\frac{1}{2}\big)-7\\ &=4\big(\frac{1}{4}\big) - \frac{4}{2} -7\\ &=\frac{4}{4}-2-7\\ &=1-2-7\\ &=-8. \end{align*}\]

The y-value of the vertex is \(k=-8\).

Thus the vertex is at

\[\big(-\frac{1}{2},-8\big).\]

(iv) Graph of parabola using three points

The x-intercepts \((x_1,0)\) and \((x_2,0)\) are

\[\big(-\frac{1}{2}-\sqrt{2},0\big),\quad \big(-\frac{1}{2}+\sqrt{2},0\big)\]

and the vertex \((h,k)\) is

\[\big(-\frac{1}{2},-8\big).\]