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3 years ago

7

Vertex(8,3); point (4,19) find the rule of a quadratic function whose graph has the given vertex and passes through the given po

int.

1 answer:

yanalaym [24]3 years ago

5 0

Y=1(x-8)² +3 or f(x)=1(x-8)² +3

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Is will like to make this as simple as i can. -1 x 372 - -31= what

mario62 [17]

The answer is -403

The order of operations puts multiplication before subtraction. -1 x 372= -372, and -372-31= -403

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3 years ago

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The formula for the equation describing a straight line is y = b0 + b1x. the value for b1 in this equation represents the ______

Rasek [7]

Equation of a straight line is normally in the form: y = mx + c.

Where, m and c are constants in which;
m = gradient
c = y-intercept.

Comparing this standard way way of writing the equation of a straight line with the current scenario, this equation can be rewritten as;
y = b1x + b0.

This way, b1 = gradient of the line while b0 = y-intercept.

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3 years ago

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Alexxx [7]

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Step-by-step explanation:

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4 years ago

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tiny-mole [99]

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3 years ago

Y=x2-2x-5<br> Change to vertex form

MaRussiya [10]

The vertex form is $y=(x-1)^{2} -6$

Step-by-step explanation:

The given equation $y=x^{2} -2x-5$ is a parabola. The vertex of the parabola is of the form $y=ax^{2} +bx+c$

The parameters of the parabola are $a=1, b=-2$ and $c=-5$

The vertex of the x-coordinate is given by $x=\frac{-b}{2a}$

Thus, substituting the values of a and b in $x=\frac{-b}{2a}$ , we get,

$\begin{aligned}x &=\frac{-(-2)}{2(1)} \\&=\frac{2}{2} \\x &=1\end{aligned}$

Now, substituting $x=1$ in $y=x^{2} -2x-5$ , we get,

$\begin{aligned}y &=1^{2}-2(1)-5 \\&=1-2-5 \\y &=-6\end{aligned}$

Thus, the vertex is $(1,-6)$

The equation of parabola to be in vertex form is $y=a(x-h)^{2} +k$

where a is the coefficient of $x^{2}$ , h is vertex of the x-coordinate and k is the vertex of the y-coordinate.

Hence, $a=1, h=1$ and $k=-6$

Thus, substituting the values in the equation of the vertex form,

$y=1(x-1)^{2} -6\\y=(x-1)^{2} -6$

Hence, the vertex form is $y=(x-1)^{2} -6$

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4 years ago