# Thread: quadradic function

1. ## quadradic function

1a)

The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

1b)

Determine the turning point on the curve and find whether it is a maximum or a minimum.

1c)

Draw a graph of S=3T squared + 10T - 8 between T= -5 and T=3 and henc solve this equation 3T squared + 10T - 8 = 0.

2. Originally Posted by andy18
1a)

The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

1b)

Determine the turning point on the curve and find whether it is a maximum or a minimum.
Standard form of a quadratic function:

$ax^2+bx+c=0$

M=f(x)=0

$2x^2-13x+15=0$

Coefficient of $x^2$ is positive so you have a minimal value.

Vertex of the parabola is $(\frac{-b}{2a}, f(\frac{-b}{2a}))$

$\frac{-b}{2a}=\frac{13}{4}$

$f(\frac{13}{4})=2(\frac{12}{4})^2-13(\frac{13}{4})+15 = \frac{-48}{8}$

Turning point is the vertex $(\frac{13}{4}, \frac{-49}{8})$

Calculate x by factoring:
$(2x-3)(x-5)=0$
$x=\frac{3}{2} \; \; or \; \;x=5$

or use the quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

3. Originally Posted by andy18
1a)

The bending moment M for a beam at distance X metres from the support is given by the quadratic equation M = 2x squared - 13x + 15, using the quadratic formula calculate the values of X where the bending moment M=0.

1b)

Determine the turning point on the curve and find whether it is a maximum or a minimum.
I am assuming this is not a Calculus based problem.

$M = 2x^2 - 13x + 15 = 0$

So
$x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4(2)(15)}}{2 \cdot 2}$

$x = \frac{13} \pm \sqrt{49}}{4}$

$x = \frac{13 \pm 7}{4}$

$x = \frac{5, \frac{3}{2}$

For the turning point, the axis of symmetry of the parabola is at
$x = -\frac{-13}{2 \cdot 2} = \frac{13}{4}$

So
$M = 2 \left ( \frac{13}{4} \right ) ^2 - 13 \left ( \frac{13}{4} \right ) + 15 = -\frac{49}{8}$

Since the leading coefficient (2) is positive this will be a minimum point.

-Dan

4. Originally Posted by andy18
1c)

Draw a graph of S=3T squared + 10T - 8 between T= -5 and T=3 and henc solve this equation 3T squared + 10T - 8 = 0.
Here's the graph. What do you get for the answer?

-Dan