•  A quadratic function can be expressed in the form \(f(x)=ax^2+bx+c\), where \(a\), \(b\) and \(c\) are constants and \(a \ne 0\).

• If \(a\gt 0\), graph has the shape \(\LARGE \smile\) which passes through a minimum point.

• If \(a\lt 0\), graph has the shape \(\LARGE \frown\) which passes through a maximum point.

\(b^2-4ac>0\)

\(b^2-4ac=0\)

\(b^2-4ac<0\)

• A quadratic function can be expressed in the form \(f(x)=a(x-h)^2+k\) where \(a\), \(h\) and \(k\) are constants and \(a \neq 0\).
• In this form, \(x=h\) is an axis of symmetry and \((h,k)\) is the coordinate of minimum or maximum point.

• A quadratic function can be expressed in the form \(f(x)=a(x-p)(x-q)\) where \(a\), \(p\) and \(q\) are constants and \(a \neq 0\).
• In this form, the roots of the quadratic function are \(p\) and \(q\).
• The axis of symmetry, \(x=\dfrac{p+q}{2}\).
• The coordinate of minimum or maximum point is \(\left[ \dfrac{p+q}{2}, f \left( \dfrac{p+q}{2} \right) \right]\).

Express quadratic function, \(f(x)=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8}\) in the intercept form, \(f(x)=a(x-p)(x-q)\), where \(a\), \(p\), and \(q\) are constants and \(p \lt q\). Hence, state the values of \(a\), \(p\), and \(q\).

Convert the vertex form of the quadratic function into the general form first.

\(\begin{aligned} f(x)&=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8} \\ &=2\left(x^2+\dfrac{9}{2}x+\dfrac{81}{16}\right)-\dfrac{1}{8}\\ &=2x^2+9x+10 \\ &=(2x+5)(x+2)\\ &=2\left(x+\dfrac{5}{2}\right)(x+2). \end{aligned}\)

Thus, the quadratic function in the intercept form for \(f(x)=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8}\) can be expressed as \(f(x)=2\left(x+\dfrac{5}{2}\right)(x+2)\), where \(a=2\), \(p=-\dfrac{5}{2}\), and \(q=-2\).

•  A quadratic function can be expressed in the form \(f(x)=ax^2+bx+c\), where \(a\), \(b\) and \(c\) are constants and \(a \ne 0\).

• If \(a\gt 0\), graph has the shape \(\LARGE \smile\) which passes through a minimum point.

• If \(a\lt 0\), graph has the shape \(\LARGE \frown\) which passes through a maximum point.

\(b^2-4ac>0\)

\(b^2-4ac=0\)

\(b^2-4ac<0\)

• A quadratic function can be expressed in the form \(f(x)=a(x-h)^2+k\) where \(a\), \(h\) and \(k\) are constants and \(a \neq 0\).
• In this form, \(x=h\) is an axis of symmetry and \((h,k)\) is the coordinate of minimum or maximum point.

• A quadratic function can be expressed in the form \(f(x)=a(x-p)(x-q)\) where \(a\), \(p\) and \(q\) are constants and \(a \neq 0\).
• In this form, the roots of the quadratic function are \(p\) and \(q\).
• The axis of symmetry, \(x=\dfrac{p+q}{2}\).
• The coordinate of minimum or maximum point is \(\left[ \dfrac{p+q}{2}, f \left( \dfrac{p+q}{2} \right) \right]\).

Express quadratic function, \(f(x)=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8}\) in the intercept form, \(f(x)=a(x-p)(x-q)\), where \(a\), \(p\), and \(q\) are constants and \(p \lt q\). Hence, state the values of \(a\), \(p\), and \(q\).

Convert the vertex form of the quadratic function into the general form first.

\(\begin{aligned} f(x)&=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8} \\ &=2\left(x^2+\dfrac{9}{2}x+\dfrac{81}{16}\right)-\dfrac{1}{8}\\ &=2x^2+9x+10 \\ &=(2x+5)(x+2)\\ &=2\left(x+\dfrac{5}{2}\right)(x+2). \end{aligned}\)

Thus, the quadratic function in the intercept form for \(f(x)=2\left(x+\dfrac{9}{4}\right)^2-\dfrac{1}{8}\) can be expressed as \(f(x)=2\left(x+\dfrac{5}{2}\right)(x+2)\), where \(a=2\), \(p=-\dfrac{5}{2}\), and \(q=-2\).