In short, for a vertical parabola, namely one whose independent variable is on the x-axis, usually is x², if the leading term coefficient is negative, the parabola opens downward, and its peak or vertex is at a maximum, check the picture below at the left-hand-side.

and when the leading term coefficient is positive, the parabola opens upwards, with a minimum, check the picture below at the right-hand-side.

8 0

3 years ago

Suppose Triangle ABC has vertices A(-2,10), B(-4,5) and C(3,2). It is dilated with a scale factor of ¼. What are the vertices of

stiv31 [10]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

M+2/3=1/4m-1 what the formula used to find the value of m?

zimovet [89]

Answer:

The value of m is $\frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6}$ by using quadratic formula

Solution:

Given, expression is $m+\frac{2}{3}=\frac{1}{4 m}-1$

Now, we have to solve the above given expression.

$\text { Now, } \mathrm{m}+\frac{2}{3}=\frac{1}{4 m}-1$

By multiplying the equation with m, we get

$\begin{array}{l}{m^{2}+\frac{2}{3} m+m=\frac{1}{4}} \\\\ {m^{2}+m\left(\frac{2}{3}+1\right)=\frac{1}{4}} \\\\ {m^{2}+\frac{5}{3} m=\frac{1}{4}}\end{array}$

$\begin{array}{l}{12 m^{2}+20 m=3} \\ {12 m^{2}+20 m-3=0}\end{array}$

Now, let us use quadratic formula

$\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in our problem, a = 12, b = 20, c = -3

$\begin{array}{l}{m=\frac{-20 \pm \sqrt{20^{2}-4 \times 12 \times(-3)}}{2 \times 12}} \\\\ {=\frac{-20 \pm \sqrt{400+144}}{24}} \\\\ {=\frac{-20 \pm \sqrt{544}}{24}} \\\\ {=\frac{-20 \pm 4 \sqrt{34}}{24}=\frac{-5 \pm \sqrt{34}}{6}} \\\\ {=\frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6}}\end{array}$

Hence the value of m is $\frac{-5+\sqrt{34}}{6} \text { or } \frac{-5-\sqrt{34}}{6}$ by using quadratic formula

7 0

3 years ago