Question

Please help me solve these are quadratic equations: I​

Answer 1

Answer:

$\boxed {\boxed {\sf a= 3, b=7, c=2}}$

$\boxed {\boxed {\sf Y-intercept = c }}$

$\boxed {\boxed {\sf Minimum}}$

Step-by-step explanation:

a. a, b, c

Quadratic equations in standard form are written as:

$f(x)= ax^2+bx+c$

The exponents are listed from greatest to least. The quadratic is already in this form, so match up the variables and values in the equation.

$f(x)= 3x^2+7x+2$

$a=3 \\b=7 \\c=2$

b. Y-intercept

The y-intercept for a quadratic is c.

You can also substitute 0 in for x, because the y-intercept occurs when the quadratic crosses the y-axis (at x=0).

$f(0)= 3(0)^2+7(0)+2$

$f(0)=3(0)+7(0)+2\\f(0)= 0+0+2\\f(0)=2$

We know that c=2, so the y-intercept is c.

c. Maximum/minimum

The entire quadratic is positive, so the quadratic opens up and has a minimum.